Q95 / 180 · COMEDK 2025 (25 May Shift) · Mathematics

$\int \dfrac{2 x}{\left(x^2+1\right)\left(x^2+2\right)^2} d x=$

  • A
    $\log \left|x^2+1\right|+\log \left|x^2+2\right|+\dfrac{1}{x^2+2}+c$
  • B
    $\log \left|x^2+1\right|-\log \left|x^2+2\right|+\dfrac{1}{x+1}+c$
  • C
    $\log \left|\dfrac{x^2+1}{x^2+2}\right|+\tan ^{-1} x+c$
  • Correct answer
    D
    $\log \left|\dfrac{x^2+1}{x^2+2}\right|+\dfrac{1}{x^2+2}+c$

Solution

Let $I = \int \dfrac{2x}{(x^2+1)(x^2+2)^2} dx$.

Substitute $t = x^2$, then $dt = 2x dx$.

The integral becomes $I = \int \dfrac{dt}{(t+1)(t+2)^2}$.

Using partial fractions, let $\dfrac{1}{(t+1)(t+2)^2} = \dfrac{A}{t+1} + \dfrac{B}{t+2} + \dfrac{C}{(t+2)^2}$.

Multiplying by $(t+1)(t+2)^2$, we get $1 = A(t+2)^2 + B(t+1)(t+2) + C(t+1)$.

For $t = -1$, $1 = A(1)^2 \Rightarrow A = 1$.

For $t = -2$, $1 = C(-1) \Rightarrow C = -1$.

Comparing coefficients of $t^2$, $0 = A + B \Rightarrow B = -A = -1$.

Thus, $I = \int \left( \dfrac{1}{t+1} - \dfrac{1}{t+2} - \dfrac{1}{(t+2)^2} \right) dt$.

Integrating, $I = \ln|t+1| - \ln|t+2| - \left( -\dfrac{1}{t+2} \right) + c = \ln\left| \dfrac{t+1}{t+2} \right| + \dfrac{1}{t+2} + c$.

Substituting $t = x^2$ back, $I = \ln\left| \dfrac{x^2+1}{x^2+2} \right| + \dfrac{1}{x^2+2} + c$.

Answer: $\log \left|\dfrac{x^2+1}{x^2+2}\right|+\dfrac{1}{x^2+2}+c$

Hence, the correct answer is option: 'D'.

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