If I_n = int (log x)^n , dx,then I_n + n I_{n | vedclass

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

Question

(A) Given $I_n = \int (\log x)^n \, dx$. Using integration by parts,let $u = (\log x)^n$ and $dv = dx$. Then $du = n(\log x)^{n-1} \cdot \frac{1}{x} \

Vedclass · Accepted Answer

$x(\log x)^n$

Vedclass · Answer

(A) Given $I_n = \int (\log x)^n \, dx$. Using integration by parts,let $u = (\log x)^n$ and $dv = dx$. Then $du = n(\log x)^{n-1} \cdot \frac{1}{x} \, dx$ and $v = x$. Applying the formula $\int u \, dv = uv - \int v \, du$: $I_n = x(\log x)^n - \int x \cdot n(\log x)^{n-1} \cdot \frac{1}{x} \, dx$ $I_n = x(\log x)^n - n \int (\log x)^{n-1} \, dx$ Since $I_{n-1} = \int (\log x)^{n-1} \, dx$,we have: $I_n = x(\log x)^n - n I_{n-1}$ Rearranging the terms,we get: $I_n + n I_{n-1} = x(\log x)^n$.

If $I_n = \int (\log x)^n \, dx$,then $I_n + n I_{n-1} =$

Similar Questions

If $\int \sin ^{-1}\left(\frac{2 x}{1+x^2}\right) d x=f(x)-\log \left(1+x^2\right)$, then $f(x)$ is equal to

The integral $\int {x\,{{\cos }^{ - 1}}\,\left( {\frac{{1 - {x^2}}}{{1 + {x^2}}}} \right)dx} \,\left( {x > 0} \right)$ is equal to

$\int \frac{\log x}{x^3} \, dx = $

$\int (\log x)^2 dx =$

Prove that $\int_{0}^{1} x e^{x} d x = 1$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module