$\int(\log x)^3 x^5 d x=$

$\int(\log x)^3 x^5 d x=$
  1. $x^6\left[\frac{(\log x)^3}{12}-\frac{1}{6}(\log x)^2+\frac{1}{6} \log x-\frac{1}{36}\right]+c$
  2. $x^6\left[\frac{(\log x)^3}{6}-\frac{1}{18}(\log x)^2+\frac{\log x}{12}-\frac{1}{36}\right]+c$
  3. $x^6\left[\frac{(\log x)^3}{6}+\frac{1}{12}(\log x)^2-\frac{\log x}{12}+\frac{1}{36}\right]+c$
  4. $x^6\left[\frac{(\log x)^3}{6}-\frac{(\log x)^2}{12}+\frac{\log x}{36}-\frac{1}{216}\right]+c$

Solution

$\int(\log x)^3 x_{\text {II }}^5 d x$ Using by parts $ \begin{aligned} & =(\log x)^3 \int x^5 d x-\int\left[\frac{d}{d x}(\log x)^3 \int x^5 d x\right] d x \\ & =(\log x)^3 \cdot \frac{x^6}{6}-\int \frac{3 \cdot(\log x)^2}{x} \cdot \frac{x^6}{6} d x \\ & =\frac{x^6}{6}(\log x)^3-\frac{1}{2} \int \underset{\mathrm{I}}{\operatorname{ta}}(\log x)^2 \cdot x^5 d x \\ & =\frac{x^6}{6}(\log x)^3-\frac{1}{2} \\ & =\frac{x^6}{6}(\log x)^3-\frac{1}{12} x^6(\log x)^2+\frac{1}{2} \int \frac{2 \log x}{x} \cdot \frac{x^6}{6} d x \\ & =\frac{x^6}{6}(\log x)^3-\frac{x^6}{12}(\log x)^2+\frac{1}{6} \int(\log x) \cdot x^5 d x \\ & =\frac{x^6}{6}(\log x)^3-\frac{x^6}{12}(\log x)^2 \\ & \quad+\frac{1}{6}\left[\log x \int x^5 d x-\int x-\int\left\{\frac{d}{d x}(\log x) \int x^5 d x\right\} d x\right. \end{aligned} $ $\begin{aligned} & =\frac{x^6}{6}(\log x)^3-\frac{x^6}{12}(\log x)^2+\frac{1}{6}(\log x) \cdot \frac{x^6}{6} \\ & =\frac{x^6}{6}(\log x)^3-\frac{x^6}{12}(\log x)^2+\frac{x^6}{36} \cdot \frac{x^6}{6} d x \\ & =\frac{x^6}{6}\left(\log x-\frac{1}{36} \int x^5 d x\right. \\ & =x^6\left[\frac{x^6}{12}(\log x)^2+\frac{x^6}{36}(\log x)-\frac{1}{36}\left(\frac{x^6}{6}\right)+C\right. \\ \end{aligned}$

Asked in: AP EAMCET 2018 (24 Apr Shift 1)

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