Q112 / 160 · TS EAMCET 2018 (05 May Shift 2) · Mathematics

If $\int \frac{x^4+1}{x^6+1} d x=A \tan ^{-1} x+B \tan ^{-1} x^3+c$, then $(A, B)=$

  • Correct answer
    A
    $\left(1, \frac{1}{3}\right)$
  • B
    $\left(1, \frac{1}{4}\right)$
  • C
    $\left(1, \frac{1}{6}\right)$
  • D
    $\left(1, \frac{4}{3}\right)$

Solution

We have, $$ \int \frac{x^4+1}{x^6+1} d x=A \tan ^{-1} x+B \tan ^{-1} x^3+c $$ Let $I=\int \frac{x^4+1}{x^6+1} d x=\int \frac{x^4+1+x^2-x^2}{\left(x^2\right)^3+(1)^3} d x$ $$ \begin{aligned} = & \int \frac{\left(x^4-x^2+1\right)+x^2}{\left(x^4+1\right)\left(x^4+1-x^2\right)} d x \\ & {\left[\because a^3+b^3=(a+b)\left(a^2+b^2-a b\right)\right] } \\ = & \int \frac{d x}{x^2+1}+\int \frac{x^2}{\left(x^2+1\right)\left(x^4+1-x^2\right)} d x \\ = & \int \frac{d x}{x^2+1}+\int \frac{x^2}{x^6+1} d x \\ = & \tan ^{-1} x+\int \frac{x^2}{\left(x^3\right)^2+1} d x \end{aligned} $$ Put $x^3=t$ $$ \begin{aligned} \Rightarrow \quad 3 x^2 d x & =d t \Rightarrow x^2 d x=\frac{1}{3} d t \\ \therefore \quad \quad \quad & =\tan ^{-1} x+\frac{1}{3} \int \frac{d t}{t^2+1} \\ & =\tan ^{-2} x+\frac{1}{3} \tan ^{-1}(t)+c \\ & =\tan ^{-2} x+\frac{1}{3} \tan ^{-1}\left(x^3\right)+c \end{aligned} $$ On comparing it with Eq.(i), we get $A=1$ and $B=\frac{1}{3}$.

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

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