int (log x)^3 x^5 dx = | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $I = \int (\log x)^3 x^5 dx$. Using integration by parts $\int u v dx = u \int v dx - \int (u' \int v dx) dx$,where $u = (\log x)^3$ and $v =

Vedclass · Accepted Answer

$x^6 \left[ \frac{(\log x)^3}{6} - \frac{(\log x)^2}{12} + \frac{\log x}{36} - \frac{1}{216} \right] + c$

Vedclass · Answer

(A) Let $I = \int (\log x)^3 x^5 dx$. Using integration by parts $\int u v dx = u \int v dx - \int (u' \int v dx) dx$,where $u = (\log x)^3$ and $v = x^5$. $I = (\log x)^3 \frac{x^6}{6} - \int 3(\log x)^2 \frac{1}{x} \frac{x^6}{6} dx = \frac{x^6}{6}(\log x)^3 - \frac{1}{2} \int x^5 (\log x)^2 dx$. Applying parts again for $\int x^5 (\log x)^2 dx$: $I = \frac{x^6}{6}(\log x)^3 - \frac{1}{2} \left[ (\log x)^2 \frac{x^6}{6} - \int 2(\log x) \frac{1}{x} \frac{x^6}{6} dx \right] = \frac{x^6}{6}(\log x)^3 - \frac{x^6}{12}(\log x)^2 + \frac{1}{6} \int x^5 \log x dx$. Applying parts again for $\int x^5 \log x dx$: $I = \frac{x^6}{6}(\log x)^3 - \frac{x^6}{12}(\log x)^2 + \frac{1}{6} \left[ \log x \frac{x^6}{6} - \int \frac{1}{x} \frac{x^6}{6} dx \right] = \frac{x^6}{6}(\log x)^3 - \frac{x^6}{12}(\log x)^2 + \frac{x^6}{36} \log x - \frac{1}{36} \int x^5 dx$. $I = \frac{x^6}{6}(\log x)^3 - \frac{x^6}{12}(\log x)^2 + \frac{x^6}{36} \log x - \frac{1}{36} \frac{x^6}{6} + c = x^6 \left[ \frac{(\log x)^3}{6} - \frac{(\log x)^2}{12} + \frac{\log x}{36} - \frac{1}{216} \right] + c$.

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

Similar Questions

$\int e^{\sqrt{x}} \, dx$ is equal to ($A$ is an arbitrary constant).

$\int x{{\sin }^{ - 1}}x\;dx = $

$\int \sqrt{x} e^{\sqrt{x}} \, dx = $

If $I_n = \int x^n \sin x \, dx$ and $I_6 - 360 I_2 = f(x) \cos x + g(x) \sin x$,then $f(1) + g(1) =$

Integrate the function: $\left(x^{2}+1\right) \log x$

Vedclass Test Series

Exam Paper Generator

Online Exam Module