int {{x^3}log x,dx = } | vedclass

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

Question

(B) To solve the integral $I = \int {{x^3}\log x\,dx}$,we use the method of integration by parts: $\int {u\,dv} = uv - \int {v\,du}$.Let $u = \log x$

Vedclass · Accepted Answer

$\frac{1}{{16}}[4{x^4}\log x - {x^4}] + c$

Vedclass · Answer

(B) To solve the integral $I = \int {{x^3}\log x\,dx}$,we use the method of integration by parts: $\int {u\,dv} = uv - \int {v\,du}$.Let $u = \log x$ and $dv = x^3\,dx$.Then $du = \frac{1}{x}\,dx$ and $v = \frac{x^4}{4}$.Applying the formula:$I = \left( \log x \cdot \frac{x^4}{4} \right) - \int {\frac{x^4}{4} \cdot \frac{1}{x}\,dx} + c$$I = \frac{x^4}{4}\log x - \frac{1}{4}\int {x^3\,dx} + c$$I = \frac{x^4}{4}\log x - \frac{1}{4} \cdot \frac{x^4}{4} + c$$I = \frac{x^4}{4}\log x - \frac{x^4}{16} + c$Taking $\frac{1}{16}$ as a common factor:$I = \frac{1}{16}[4x^4\log x - x^4] + c$.

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

Similar Questions

If $\int \cos x \log \left(\tan \frac{x}{2}\right) dx = \sin x \log \left(\tan \frac{x}{2}\right) + f(x)$, then $f(x)$ is equal to (assuming $c$ is an arbitrary real constant).

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

$\int e^{x} \cdot x^{5} \, dx$ is

$\int {\frac{{{{\cot }^{ - 1}}({e^x})}}{{{e^x}}}} dx$ is equal to -

$\int \cos (\log x) d x=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module