Q113 / 160 · AP EAMCET 2022 (08 Jul Shift 2) · Mathematics
$\int(\log x)^2 x^3 d x=\frac{x^4}{32} f(x)+C \Rightarrow f(x)=$
- Correct answerA$8(\log x)^2-4 \log x+1$
- B$8 \log x-4 x^4+x^3$
- C$8(\log x)^2+4 x-x^2$
- D$4(\log x)^2-4 x^2+x+1$
Solution
$I=\int(\log x)^2 x^3 d x$
Using integration by parts,
$I=(\log x)^2 \int x^3 d x-\int\left(\frac{d(\log x)^2}{d x} \int x^3 d x\right) d x$
$\begin{aligned} & =\frac{(\log x)^2 x^4}{4}-\int \frac{2 \log x}{x} \times \frac{x^4}{4} d x \\ & \quad=\frac{(\log x)^2 x^4}{4}-\frac{1}{2} \int(\log x) x^3 d x \\ & \quad=\frac{(\log x)^2 x^4}{4}-\frac{1}{2}\end{aligned}$
$\left[(\log x) \int x^3 d x-\int\left(\frac{d \log x}{d x} \int x^3 d x\right) d x\right]$
$\begin{aligned} & =\frac{(\log x)^2 x^4}{4}-\frac{1}{2}\left[\frac{(\log x) x^4}{4}-\int \frac{1}{x} \times \frac{x^4}{4} d x\right] \\ & =\frac{(\log x)^2 x^4}{4}-\frac{1}{8}(\log x) x^4+\frac{1}{8} \times \frac{x^4}{4}+C \\ & =\frac{x^4}{32}\left[8(\log x)^2-4(\log x)+1\right]+C\end{aligned}$
$\therefore f(x)=8(\log x)^2-4(\log x)+1$
Hence, the correct answer is option: 'A'.