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Q11:
Choose the correct answer in Exercises 10 to 11. 11. $\int \sqrt{x^2-8x+7} dx$ is equal to

Solution :

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

x y f(x) = sqrt(x^2-8x+7)

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



CBSE Solutions for Class 12 Mathematics Integrals


Chapters in CBSE - Class 12 Mathematics


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