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Q32:
Prove the following (Exercises 32 to 37) 32. $\int_{1}^{3} \frac{dx}{x^2(x+1)} = \frac{2}{3} + \log \frac{2}{3}$

Solution :

Step 1: Decomposition via Partial Fractions

To evaluate the integral $I = \int_{1}^{3} \frac{dx}{x^2(x+1)}$, we express the integrand as a sum of partial fractions [Per the Method of Partial Fractions]:

$\frac{1}{x^2(x+1)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x+1}$

Multiplying both sides by $x^2(x+1)$, we obtain the identity:

$1 = Ax(x+1) + B(x+1) + Cx^2$

$1 = A(x^2 + x) + Bx + B + Cx^2$

$1 = (A+C)x^2 + (A+B)x + B$

Step 2: Solving for Coefficients

Equating coefficients of like powers of $x$ on both sides:

  • Constant term: $B = 1$
  • Coefficient of $x$: $A + B = 0 \implies A + 1 = 0 \implies A = -1$
  • Coefficient of $x^2$: $A + C = 0 \implies -1 + C = 0 \implies C = 1$

Thus, the integrand is: $\frac{1}{x^2(x+1)} = -\frac{1}{x} + \frac{1}{x^2} + \frac{1}{x+1}$

Step 3: Integration

We now evaluate the definite integral using the Fundamental Theorem of Calculus:

$I = \int_{1}^{3} \left( -\frac{1}{x} + \frac{1}{x^2} + \frac{1}{x+1} \right) dx$

$I = \left[ -\log|x| - \frac{1}{x} + \log|x+1| \right]_{1}^{3}$

$I = \left[ \log\left| \frac{x+1}{x} \right| - \frac{1}{x} \right]_{1}^{3}$

Step 4: Applying Limits

Substitute the upper and lower limits into the antiderivative:

$I = \left( \log\left| \frac{3+1}{3} \right| - \frac{1}{3} \right) - \left( \log\left| \frac{1+1}{1} \right| - \frac{1}{1} \right)$

$I = \left( \log\frac{4}{3} - \frac{1}{3} \right) - \left( \log 2 - 1 \right)$

$I = \log\frac{4}{3} - \log 2 - \frac{1}{3} + 1$

$I = \log\left( \frac{4/3}{2} \right) + \frac{2}{3}$

$I = \log\left( \frac{2}{3} \right) + \frac{2}{3}$

x y 1 3 Area under f(x) = 1/(x^2(x+1))

Final Solution: \frac{2}{3} + \log \frac{2}{3}


More Questions from Class 12 Mathematics Integrals Miscellaneous Exercise on Chapter 7


CBSE Solutions for Class 12 Mathematics Integrals


Chapters in CBSE - Class 12 Mathematics


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