Q113 / 160 · TS EAMCET 2020 (10 Sep Shift 2) · Mathematics
If $\int x^3(\log x)^2 d x=x^4\left[A(\log x)^2+B(\log x)\right.$ $+C \log e]+K$, then $A+B+C$
- A$\frac{7}{24}$
- B$\frac{4}{25}$
- C$\frac{3}{14}$
- Correct answerD$\frac{5}{32}$
Solution
We have,
$\int x^3(\log x)^2 d x=x^4$ $\left[A(\log x)^2+B(\log x)+C \log e\right]+K$
Now, $\int x^3(\log x)^2 d x$
$=(\log x)^2 \int x^3 d x-\int\left(\frac{d}{d x}(\log x)^2 \int x^3 d x\right) d x$
$\left.=(\log x)^2 \cdot \frac{x^4}{4}-\int(2 \log x) \cdot \frac{1}{x} \cdot \frac{x^4}{4}\right) d x$
$=(\log x)^2\left(\frac{x^4}{4}\right)-\frac{1}{2} \int \log x \cdot x^3 d x$
$=(\log x)^2\left(\frac{x^4}{4}\right)-\frac{1}{2}$
$\left[\log x \int x^3 d x-\int\left(\frac{d}{d x}\left(\log x \int x^3 d x\right) d x\right]\right.$
$=(\log x)^2\left(\frac{x^4}{4}\right)-\frac{1}{2}\left[\log x\left(\frac{x^4}{4}\right)-\int \frac{1}{x} \cdot \frac{x^4}{4} d x\right]$
$=(\log x)^2\left(\frac{x^4}{4}\right)-\frac{1}{2}\left[\log x\left(\frac{x^4}{4}\right)-\frac{1}{4} \cdot \int x^3 d x\right]$
$=(\log x)^2\left(\frac{x^4}{4}\right)-\frac{1}{2}\left[\log x\left(\frac{x^4}{4}\right)-\frac{1}{4} \cdot \frac{x^4}{4}\right]+K$
$=x^4\left[\frac{1}{4}(\log x)^2-\frac{1}{8} \log x+\frac{1}{32} \log e\right]+K$
$\therefore A=\frac{1}{4}, B=\frac{-1}{8}$ and $C=\frac{1}{32}$
$\therefore A+B+C=\frac{1}{4}-\frac{1}{8}+\frac{1}{32}=\frac{8-4+1}{32}=\frac{5}{32}$
Hence, the correct answer is option: 'D'.