$\int e^{3 \log x}\left(x^4+1\right)^{-1} d x=$

$\int e^{3 \log x}\left(x^4+1\right)^{-1} d x=$
  1. $e^{3 \log x}+c$
  2. $\frac{1}{4} \log \left(x^4+1\right)+c$
  3. $\frac{1}{3} \log \left(x^4+1\right)+c$
  4. $\frac{x^4}{x^4+1}$

Solution

$I=\int e^{3 \log x}\left(x^4+1\right)^{-1} d x=\int x^3\left(x^4+1\right)^{-1} d x$ Let $x^4+1=t \Rightarrow 4 x^3 d x=d t$ $ \therefore \quad I=\frac{1}{4} \int \frac{d t}{t}=\frac{1}{4} \log _e(t)+C=\frac{1}{4} \log _e\left(x^4+1\right)+C $

Asked in: AP EAMCET 2020 (22 Sep Shift 1)

Practice more Indefinite Integration questions on Aicharya