$\smallint \,\,{x^3}\,\,\log \,\,x\,dx$ is equal to

$\frac{{{x^4}\,\,\log \,\,x}}{4}\,\, + \,\,c$

$\frac{1}{8}\,\,[{x^4}\,\,\log \,\,x\,\, - \,\,4{x^2}] + c$

JEE MATHEMATICS MCQ: INDEFINITE INTEGRATION

1 Medium

1 of 1

Application +4 | -1

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

is equal to

View answer and explanation

Correct option: B

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

Option B (

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

) is correct because it is the result of integrating the given expression using integration by parts.

Practice All Chapter MCQs →

\int \sin ^5 x \cos ^4 x d x=

f(x)

\int\limits_{0}^{t^2} x f(x) d x=\frac{2}{5} t^5

f\left(\frac{4}{25}\right)

\int \frac{1-7 \cos ^2 x}{\sin ^7 x \cos ^2 x} d x=\frac{f(x)}{\sin ^7 x}+C

\int {e}^{{x}^2}\left(2-\frac{1}{{x}^2}\right) {de}={e}^{{x}^2} {f}({x})+{C}

{f}\left(\frac{1}{2}\right)=2

f(1)

{C}

\int \frac{x^7}{\left(1-x^2\right)^5} d x

\int \,\frac{{a{x^3} + \,\,b{x^2} + c}}{{{x^4}}}\;\,dx