Step 1: Transforming the Integral into Standard Form

To evaluate the integral $I = \int \sqrt{1 - 4x^2} \, dx$, we first manipulate the radicand to fit the standard integral form $\int \sqrt{a^2 - u^2} \, du$.

Factor out $4$ from the expression inside the square root:

$I = \int \sqrt{4\left(\frac{1}{4} - x^2\right)} \, dx$

$I = 2 \int \sqrt{\left(\frac{1}{2}\right)^2 - x^2} \, dx$ [By the property of radicals $\sqrt{ab} = \sqrt{a}\sqrt{b}$]

Step 2: Applying the Standard Integral Formula

We utilize the standard integration formula for the form $\int \sqrt{a^2 - x^2} \, dx$:

$\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1}\left(\frac{x}{a}\right) + C$

Here, $a = \frac{1}{2}$ and $a^2 = \frac{1}{4}$.

Step 3: Execution of the Integration

Substitute $a = \frac{1}{2}$ into the formula:

$I = 2 \left[ \frac{x}{2} \sqrt{\left(\frac{1}{2}\right)^2 - x^2} + \frac{(1/2)^2}{2} \sin^{-1}\left(\frac{x}{1/2}\right) \right] + C$

$I = 2 \left[ \frac{x}{2} \sqrt{\frac{1}{4} - x^2} + \frac{1/4}{2} \sin^{-1}(2x) \right] + C$

$I = 2 \left[ \frac{x}{2} \sqrt{\frac{1-4x^2}{4}} + \frac{1}{8} \sin^{-1}(2x) \right] + C$

Step 4: Simplifying the Expression

Distribute the factor of $2$ into the brackets and simplify the square root term:

$I = 2 \left[ \frac{x}{2} \cdot \frac{\sqrt{1-4x^2}}{2} + \frac{1}{8} \sin^{-1}(2x) \right] + C$

$I = \frac{2x}{4} \sqrt{1-4x^2} + \frac{2}{8} \sin^{-1}(2x) + C$

$I = \frac{x}{2} \sqrt{1-4x^2} + \frac{1}{4} \sin^{-1}(2x) + C$

Geometric Representation of the Radicand Function

Final Solution: $\frac{x}{2} \sqrt{1-4x^2} + \frac{1}{4} \sin^{-1}(2x) + C$

Step 1: Completing the Square for the Quadratic Form

The integral to evaluate is $I = \int \sqrt{x^2 + 4x + 6} \, dx$. To simplify the radicand, we complete the square for the quadratic expression $x^2 + 4x + 6$.

We write the expression as:

$x^2 + 4x + 6 = (x^2 + 4x + 4) - 4 + 6$

[By the method of completing the square: $(x+a)^2 = x^2 + 2ax + a^2$ where $2a = 4 \implies a=2$]

$x^2 + 4x + 6 = (x+2)^2 + 2$ $x^2 + 4x + 6 = (x+2)^2 + (\sqrt{2})^2$

Step 2: Selecting the Standard Integration Formula

The integral now takes the form $\int \sqrt{u^2 + a^2} \, du$, where $u = (x+2)$ and $a = \sqrt{2}$. The standard integration formula for this form is:

$\int \sqrt{x^2 + a^2} \, dx = \frac{x}{2} \sqrt{x^2 + a^2} + \frac{a^2}{2} \ln|x + \sqrt{x^2 + a^2}| + C$

[Reference: Standard Integral Formula for $\sqrt{x^2 + a^2}$]

Step 3: Substitution and Integration

Applying the formula with $u = (x+2)$ and $a^2 = 2$:

$I = \int \sqrt{(x+2)^2 + (\sqrt{2})^2} \, dx$

Following the formula:

$I = \frac{(x+2)}{2} \sqrt{(x+2)^2 + 2} + \frac{2}{2} \ln|(x+2) + \sqrt{(x+2)^2 + 2}| + C$

Step 4: Simplification of the Result

Substitute $(x+2)^2 + 2$ back with the original expression $x^2 + 4x + 6$:

$I = \frac{x+2}{2} \sqrt{x^2 + 4x + 6} + 1 \cdot \ln|(x+2) + \sqrt{x^2 + 4x + 6}| + C$

Final Solution: $\frac{(x+2)}{2} \sqrt{x^2 + 4x + 6} + \ln|x + 2 + \sqrt{x^2 + 4x + 6}| + C$

Step 1: Identifying the Form

The integral to be solved is $I = \int \sqrt{1 + 3x - x^2} \, dx$. This expression is of the form $\int \sqrt{a^2 - x^2} \, dx$, where the quadratic expression under the radical requires completion of the square [Completing the Square Method].

Step 2: Completing the Square

Rewrite the expression $1 + 3x - x^2$:

$1 + 3x - x^2 = -(x^2 - 3x - 1)$

To complete the square for $x^2 - 3x$, we add and subtract $(\frac{3}{2})^2 = \frac{9}{4}$:

$= -[x^2 - 3x + \frac{9}{4} - \frac{9}{4} - 1]$

$= -[(x - \frac{3}{2})^2 - \frac{13}{4}]$

$= \frac{13}{4} - (x - \frac{3}{2})^2$

Thus, the integral becomes: $I = \int \sqrt{(\frac{\sqrt{13}}{2})^2 - (x - \frac{3}{2})^2} \, dx$

Step 3: Applying the Standard Integral Formula

We utilize the standard result: [$\int \sqrt{a^2 - u^2} \, du = \frac{u}{2} \sqrt{a^2 - u^2} + \frac{a^2}{2} \sin^{-1}(\frac{u}{a}) + C$].

Here, $a = \frac{\sqrt{13}}{2}$ and $u = (x - \frac{3}{2})$.

$I = \frac{x - \frac{3}{2}}{2} \sqrt{\frac{13}{4} - (x - \frac{3}{2})^2} + \frac{\frac{13}{4}}{2} \sin^{-1}\left(\frac{x - \frac{3}{2}}{\frac{\sqrt{13}}{2}}\right) + C$

Step 4: Algebraic Simplification

Simplify the term $(x - \frac{3}{2})$ as $(\frac{2x - 3}{2})$:

$I = \frac{2x - 3}{4} \sqrt{1 + 3x - x^2} + \frac{13}{8} \sin^{-1}\left(\frac{2x - 3}{\sqrt{13}}\right) + C$

Step 5: Final Result

Combining the terms into the final analytical form:

Final Solution: $\frac{2x-3}{4}\sqrt{1+3x-x^2} + \frac{13}{8}\sin^{-1}\left(\frac{2x-3}{\sqrt{13}}\right) + C$

Step 1: Establishing the Standard Integral Form

The given function is $\int \sqrt{4 - x^2} \, dx$. This expression corresponds to the standard integral form:

$$\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) + C$$

[Justification: This standard formula is derived via trigonometric substitution $x = a\sin\theta$ and the identity $\cos^2\theta = 1 - \sin^2\theta$.]

Step 2: Identifying Variables

We rewrite the given integrand $\sqrt{4 - x^2}$ to match the form $\sqrt{a^2 - x^2}$:

• $a^2 = 4 \implies a = 2$
• Variable of integration = $x$

Step 3: Applying the Substitution and Formula

Substituting $a = 2$ into the standard formula derived from integration by parts:

$$\int \sqrt{2^2 - x^2} \, dx = \frac{x}{2}\sqrt{2^2 - x^2} + \frac{2^2}{2}\sin^{-1}\left(\frac{x}{2}\right) + C$$

Step 4: Algebraic Simplification

Simplifying the coefficients and the constant term:

$$\frac{x}{2}\sqrt{4 - x^2} + \frac{4}{2}\sin^{-1}\left(\frac{x}{2}\right) + C$$

$$\frac{x}{2}\sqrt{4 - x^2} + 2\sin^{-1}\left(\frac{x}{2}\right) + C$$

Geometric Representation of the Area

The integral represents the area under the curve of a semi-circle with radius $r=2$ centered at the origin, constrained by the upper half of $x^2 + y^2 = 4$.

Final Solution: $\frac{x}{2}\sqrt{4 - x^2} + 2\sin^{-1}\left(\frac{x}{2}\right) + C$

Step 1: Establishing the Standard Integration Formula

To evaluate the integral $I = \int \sqrt{1+x^2} \, dx$, we utilize the standard algebraic form for the integral of the square root of a quadratic expression, specifically $\int \sqrt{x^2 + a^2} \, dx$.

The standard integration formula derived via integration by parts is:

$\int \sqrt{x^2 + a^2} \, dx = \frac{x}{2} \sqrt{x^2 + a^2} + \frac{a^2}{2} \ln \left| x + \sqrt{x^2 + a^2} \right| + C$ [By the method of Integration by Parts, where $\int u \, dv = uv - \int v \, du$]

Step 2: Identifying Given Parameters

Given the integral $\int \sqrt{1+x^2} \, dx$, we compare this to the standard form $\int \sqrt{x^2 + a^2} \, dx$:

• $a^2 = 1 \implies a = 1$

Step 3: Substitution and Calculation

Substituting $a = 1$ into the established formula:

$I = \frac{x}{2} \sqrt{x^2 + 1^2} + \frac{1^2}{2} \ln \left| x + \sqrt{x^2 + 1^2} \right| + C$

Simplifying the constants and radicands:

$I = \frac{x}{2} \sqrt{x^2 + 1} + \frac{1}{2} \ln \left| x + \sqrt{x^2 + 1} \right| + C$

Visual Representation of the Integration Function

Step 4: Analytical Verification

To verify, differentiate the resulting expression $f(x) = \frac{x}{2} \sqrt{x^2 + 1} + \frac{1}{2} \ln |x + \sqrt{x^2+1}|$:

$\frac{d}{dx} \left[ \frac{x}{2} \sqrt{x^2+1} + \frac{1}{2} \ln |x + \sqrt{x^2+1}| \right]$

$= \left( \frac{1}{2} \sqrt{x^2+1} + \frac{x}{2} \cdot \frac{2x}{2\sqrt{x^2+1}} \right) + \frac{1}{2} \left( \frac{1 + \frac{x}{\sqrt{x^2+1}}}{x + \sqrt{x^2+1}} \right)$

$= \frac{x^2+1 + x^2}{2\sqrt{x^2+1}} + \frac{1}{2} \left( \frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}(x+\sqrt{x^2+1})} \right) = \frac{2x^2+1}{2\sqrt{x^2+1}} + \frac{1}{2\sqrt{x^2+1}} = \frac{2x^2+2}{2\sqrt{x^2+1}} = \sqrt{x^2+1}$

[Verification confirms the derivative returns the original integrand]

Final Solution: $\frac{2}{x} \sqrt{x^2+1} + \frac{1}{2} \ln |x + \sqrt{x^2+1}| + C$

Step 1: Completing the Square for the Quadratic Expression

To integrate the expression $\int \sqrt{x^2+4x+1} \, dx$, we first complete the square for the radicand $x^2 + 4x + 1$.

The general form $x^2 + bx$ becomes $(x + \frac{b}{2})^2 - (\frac{b}{2})^2$. Here, $b = 4$.

• Coefficient of $x$ is $4$. Half of $4$ is $2$.
• $x^2 + 4x + 1 = (x^2 + 4x + 4) - 4 + 1$ [Completing the Square Method]
• $x^2 + 4x + 1 = (x + 2)^2 - 3$
• $x^2 + 4x + 1 = (x + 2)^2 - (\sqrt{3})^2$

Step 2: Applying the Standard Integration Formula

We utilize the standard integral formula for the form $\int \sqrt{x^2 - a^2} \, dx$:

$\int \sqrt{x^2 - a^2} \, dx = \frac{x}{2} \sqrt{x^2 - a^2} - \frac{a^2}{2} \ln|x + \sqrt{x^2 - a^2}| + C$ [Per standard integration identity for square roots]

Step 3: Substitution and Calculation

Let $u = x + 2$, then $du = dx$. The integral becomes:

$\int \sqrt{u^2 - (\sqrt{3})^2} \, du$

Substituting $u = x + 2$ and $a = \sqrt{3}$ into the formula:

$= \frac{(x+2)}{2} \sqrt{(x+2)^2 - (\sqrt{3})^2} - \frac{(\sqrt{3})^2}{2} \ln|(x+2) + \sqrt{(x+2)^2 - (\sqrt{3})^2}| + C$

Step 4: Simplifying the Expression

Expanding the terms and reverting the radicand to its original form:

$= \frac{x+2}{2} \sqrt{x^2+4x+4-3} - \frac{3}{2} \ln|(x+2) + \sqrt{x^2+4x+1}| + C$

$= \frac{x+2}{2} \sqrt{x^2+4x+1} - \frac{3}{2} \ln|(x+2) + \sqrt{x^2+4x+1}| + C$

Geometrical Representation of the Integrand Function

Final Solution: $\frac{(x+2)}{2} \sqrt{x^2+4x+1} - \frac{3}{2} \ln |(x+2) + \sqrt{x^2+4x+1}| + C$

Step 1: Completing the Square

To integrate functions of the form $\sqrt{x^2 + bx + c}$, we must first complete the square for the quadratic expression $x^2 - 8x + 7$.

The expression is $x^2 - 8x + 7$. To complete the square, we take the coefficient of $x$ (which is $-8$), divide by $2$ to get $-4$, and square it to get $16$. We then add and subtract $16$:

$x^2 - 8x + 7 = (x^2 - 8x + 16) - 16 + 7$ [Algebraic identity: $(x-a)^2 = x^2 - 2ax + a^2$]

$= (x - 4)^2 - 9$

$= (x - 4)^2 - 3^2$

Step 2: Selecting the Integration Formula

The integral now takes the standard form $\int \sqrt{x^2 - a^2} \, dx$. The standard formula for this integral is:

$\int \sqrt{x^2 - a^2} \, dx = \frac{x}{2}\sqrt{x^2 - a^2} - \frac{a^2}{2} \ln \left| x + \sqrt{x^2 - a^2} \right| + C$ [Standard Integral Formula]

Step 3: Substitution and Evaluation

Substitute $x$ with $(x - 4)$ and $a$ with $3$ into the standard formula:

$\int \sqrt{(x - 4)^2 - 3^2} \, dx = \frac{x - 4}{2}\sqrt{(x - 4)^2 - 3^2} - \frac{3^2}{2} \ln \left| (x - 4) + \sqrt{(x - 4)^2 - 3^2} \right| + C$

Simplify the terms:

• The term $\sqrt{(x - 4)^2 - 3^2}$ is equivalent to $\sqrt{x^2 - 8x + 7}$.
• The constant $\frac{3^2}{2} = \frac{9}{2}$.

Substituting these back into the expression:

$= \frac{x - 4}{2}\sqrt{x^2 - 8x + 7} - \frac{9}{2} \ln \left| (x - 4) + \sqrt{x^2 - 8x + 7} \right| + C$

Step 4: Visualizing the Function

Final Solution: $\frac{x-4}{2}\sqrt{x^2-8x+7} - \frac{9}{2} \ln |x-4 + \sqrt{x^2-8x+7}| + C$

Step 1: Completing the Square for the Quadratic Expression

Given the integral: $I = \int \sqrt{x^2 + 3x} \, dx$.

To evaluate this, we transform the quadratic expression $x^2 + 3x$ into the form $(x+a)^2 - b^2$ [Per the method of completing the square].

Add and subtract the square of half the coefficient of $x$ (which is $3/2$):

$x^2 + 3x = x^2 + 3x + \left(\frac{3}{2}\right)^2 - \left(\frac{3}{2}\right)^2$

$= (x + \frac{3}{2})^2 - (\frac{3}{2})^2$

Step 2: Identifying the Standard Integral Form

Substitute the completed square back into the integral:

$I = \int \sqrt{(x + \frac{3}{2})^2 - (\frac{3}{2})^2} \, dx$

This follows the standard integral formula [Per the integration table for square roots of quadratics]:

$\int \sqrt{u^2 - a^2} \, du = \frac{u}{2} \sqrt{u^2 - a^2} - \frac{a^2}{2} \ln |u + \sqrt{u^2 - a^2}| + C$

Where $u = (x + \frac{3}{2})$ and $a = \frac{3}{2}$.

Step 3: Application of the Formula

Applying the variables to the formula:

$I = \frac{x + \frac{3}{2}}{2} \sqrt{(x + \frac{3}{2})^2 - (\frac{3}{2})^2} - \frac{(\frac{3}{2})^2}{2} \ln |(x + \frac{3}{2}) + \sqrt{(x + \frac{3}{2})^2 - (\frac{3}{2})^2}| + C$

Step 4: Simplifying the Expression

Simplify each component:

• The term $\frac{x + \frac{3}{2}}{2} = \frac{2x + 3}{4}$
• The term $\sqrt{(x + \frac{3}{2})^2 - (\frac{3}{2})^2} = \sqrt{x^2 + 3x}$
• The term $\frac{(\frac{3}{2})^2}{2} = \frac{9/4}{2} = \frac{9}{8}$

Substituting back into the expanded form:

$I = \frac{2x + 3}{4} \sqrt{x^2 + 3x} - \frac{9}{8} \ln |(x + \frac{3}{2}) + \sqrt{x^2 + 3x}| + C$

Final Solution: $\frac{2x + 3}{4} \sqrt{x^2 + 3x} - \frac{9}{8} \ln \left| x + \frac{3}{2} + \sqrt{x^2 + 3x} \right| + C$

Step 1: Simplify the Integral Expression

The given integral is $\int \sqrt{1 + \frac{x^2}{9}} \, dx$. First, we rewrite the expression under the square root to facilitate the application of standard integration formulas: $$\sqrt{1 + \frac{x^2}{9}} = \sqrt{\frac{9 + x^2}{9}} = \frac{1}{3} \sqrt{3^2 + x^2}$$ [By algebraic simplification of the radicand].

Step 2: Selecting the Appropriate Integration Formula

The integral now takes the form $\frac{1}{3} \int \sqrt{a^2 + x^2} \, dx$, where $a = 3$. We utilize the standard integral identity for the square root of the sum of squares: $$\int \sqrt{a^2 + x^2} \, dx = \frac{x}{2} \sqrt{a^2 + x^2} + \frac{a^2}{2} \ln \left| x + \sqrt{a^2 + x^2} \right| + C$$ [Per the integration formula for $\sqrt{a^2 + x^2}$].

Step 3: Substituting Variables into the Formula

Substitute $a = 3$ into the identity:

$$I = \frac{1}{3} \left[ \frac{x}{2} \sqrt{3^2 + x^2} + \frac{3^2}{2} \ln \left| x + \sqrt{3^2 + x^2} \right| \right] + C$$ $$I = \frac{1}{3} \left[ \frac{x}{2} \sqrt{9 + x^2} + \frac{9}{2} \ln \left| x + \sqrt{9 + x^2} \right| \right] + C$$

Step 4: Algebraic Distribution and Final Simplification

Distribute the factor $\frac{1}{3}$ across the bracketed terms:

$$I = \frac{x}{6} \sqrt{9 + x^2} + \frac{9}{6} \ln \left| x + \sqrt{9 + x^2} \right| + C$$ $$I = \frac{x}{6} \sqrt{9 + x^2} + \frac{3}{2} \ln \left| x + \sqrt{9 + x^2} \right| + C$$

To align the result with the original form $\sqrt{1 + \frac{x^2}{9}}$, note that $\sqrt{9+x^2} = \sqrt{9(1 + \frac{x^2}{9})} = 3\sqrt{1 + \frac{x^2}{9}}$:

$$I = \frac{x}{6} \cdot 3\sqrt{1 + \frac{x^2}{9}} + \frac{3}{2} \ln \left| x + 3\sqrt{1 + \frac{x^2}{9}} \right| + C$$ $$I = \frac{x}{2} \sqrt{1 + \frac{x^2}{9}} + \frac{3}{2} \ln \left| x + 3\sqrt{1 + \frac{x^2}{9}} \right| + C$$

Final Solution: \frac{x}{6} \sqrt{9+x^2} + \frac{3}{2} \ln \left| x + \sqrt{9+x^2} \right| + C

Step 1: Identifying the Form of the Integral

The integral to be solved is:

$I = \int \sqrt{x^2 + 4x - 5} \, dx$

This expression fits the standard form $\int \sqrt{x^2 + ax + b} \, dx$. To solve this, we must complete the square for the quadratic expression $x^2 + 4x - 5$.

Step 2: Completing the Square

The quadratic expression is $x^2 + 4x - 5$. Adding and subtracting $(\frac{1}{2} \text{coefficient of } x)^2 = (\frac{4}{2})^2 = 4$:

$x^2 + 4x - 5 = (x^2 + 4x + 4) - 4 - 5$

$x^2 + 4x - 5 = (x+2)^2 - 9$

$x^2 + 4x - 5 = (x+2)^2 - (3)^2$

[By Algebraic identity: $a^2 + 2ab + b^2 = (a+b)^2$]

Step 3: Applying the Standard Integral Formula

We substitute the completed square into the integral:

$I = \int \sqrt{(x+2)^2 - 3^2} \, dx$

We use the standard integration formula [Per standard integral of the form $\int \sqrt{x^2 - a^2} \, dx$]:

$\int \sqrt{x^2 - a^2} \, dx = \frac{x}{2}\sqrt{x^2 - a^2} - \frac{a^2}{2} \ln|x + \sqrt{x^2 - a^2}| + C$

In our case, substitute $x \to (x+2)$ and $a \to 3$:

$I = \frac{(x+2)}{2} \sqrt{(x+2)^2 - 3^2} - \frac{3^2}{2} \ln|(x+2) + \sqrt{(x+2)^2 - 3^2}| + C$

Step 4: Simplifying the Result

Substituting back $(x+2)^2 - 3^2 = x^2 + 4x - 5$:

$I = \frac{x+2}{2} \sqrt{x^2 + 4x - 5} - \frac{9}{2} \ln|(x+2) + \sqrt{x^2 + 4x - 5}| + C$

Final Solution: $\frac{1}{2}(x+2) \sqrt{x^2 + 4x - 5} - \frac{9}{2} \ln|x + 2 + \sqrt{x^2 + 4x - 5}| + C$

Step 1: Completing the Square for the Quadratic Expression

The integrand is given by $I = \int \sqrt{1 - 4x - x^2} \, dx$. To simplify this, we complete the square for the quadratic expression $-(x^2 + 4x - 1)$.

Consider the expression inside the square root:

$1 - 4x - x^2 = -(x^2 + 4x) + 1$

To complete the square for $(x^2 + 4x)$, we add and subtract $(\frac{4}{2})^2 = 4$:

$= -((x^2 + 4x + 4) - 4) + 1$ $= -((x+2)^2 - 4) + 1$ $= -(x+2)^2 + 4 + 1$ $= 5 - (x+2)^2$ $= (\sqrt{5})^2 - (x+2)^2$

[By the method of completing the square]

Step 2: Selecting the Integration Formula

The integral now takes the form:

$I = \int \sqrt{(\sqrt{5})^2 - (x+2)^2} \, dx$

We utilize the standard integration formula for the form $\int \sqrt{a^2 - x^2} \, dx$:

$\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) + C$

[Per the standard integral table for algebraic radicals]

Step 3: Substitution and Evaluation

Substitute $a = \sqrt{5}$ and replace $x$ with $(x+2)$ in the formula:

$I = \frac{(x+2)}{2}\sqrt{(\sqrt{5})^2 - (x+2)^2} + \frac{(\sqrt{5})^2}{2}\sin^{-1}\left(\frac{x+2}{\sqrt{5}}\right) + C$

Simplify the terms:

$I = \frac{x+2}{2}\sqrt{5 - (x^2 + 4x + 4)} + \frac{5}{2}\sin^{-1}\left(\frac{x+2}{\sqrt{5}}\right) + C$ $I = \frac{x+2}{2}\sqrt{1 - 4x - x^2} + \frac{5}{2}\sin^{-1}\left(\frac{x+2}{\sqrt{5}}\right) + C$

Visual Representation of the Integrand Function

Final Solution: $\frac{x+2}{2}\sqrt{1 - 4x - x^2} + \frac{5}{2}\sin^{-1}\left(\frac{x+2}{\sqrt{5}}\right) + C$