Question

evaluate inegral x²logxdx​

Answer 1

Correct Question :

Evaluate integral of x²logx dx.

Solution :

$\rm \longrightarrow \rm \displaystyle \int \rm {x}^{2} \: log \: x \: dx \\ \\ \\ \rm \longrightarrow \rm log \: x\displaystyle \int \rm {x}^{2} \: \: dx \: - \displaystyle \int \rm \bigg(\dfrac{d(log \: x)}{dx} \int {x}^{2} dx \bigg)dx\\ \\ \\ \rm \longrightarrow \rm( log \: x\displaystyle \rm \: \times \dfrac{ {x}^{3} }{3} \: ) \: - \displaystyle \int \rm \bigg(\frac{1}{x} \times \dfrac{ {x}^{3} }{3} \bigg)dx\\ \\ \\ \rm \longrightarrow \rm\dfrac{ {x}^{3} }{3}( log \: x\displaystyle \rm \: \: ) \: - \displaystyle \int \rm \dfrac{ {x}^{2} }{3} dx\\ \\ \\ \rm \longrightarrow \rm\dfrac{ {x}^{3} }{3}( log \: x\displaystyle \rm \: \: ) \: - \displaystyle \rm \dfrac{ {x}^{3} }{(3 \times 3)} + c\\ \\ \\ \rm \longrightarrow \rm\dfrac{ {x}^{3} }{3}( log \: x\displaystyle \rm \: \: ) \: - \displaystyle \rm \dfrac{ {x}^{3} }{9} + c$

Answer :

$\boxed{\rm\dfrac{ {x}^{3} }{3}( log \: x\displaystyle \rm \: \: ) \: - \displaystyle \rm \dfrac{ {x}^{3} }{9} + c}$

Additional Information :

$\boxed{\boxed{\begin{minipage}{4cm}\displaystyle\circ\sf\:\int{1\:dx}=x+c\\\\\circ\sf\:\int{a\:dx}=ax+c\\\\\circ\sf\:\int{x^n\:dx}=\dfrac{x^{n+1}}{n+1}+c\\\\\circ\sf\:\int{sin\:x\:dx}=-cos\:x+c\\\\\circ\sf\:\int{cos\:x\:dx}=sin\:x+c\\\\\circ\sf\:\int{sec^2x\:dx}=tan\:x+c\\\\\circ\sf\:\int{e^x\:dx}=e^x+c\end{minipage}}}$

Answer 2

Given Integrand,

$\displaystyle \sf \int {x}^{2} log(x) dx$

Integrating by parts,

$\longrightarrow \displaystyle \sf log(x) \int {x}^{2} dx - \int \bigg \{ \dfrac{d( log(x) )}{dx} \int {x}^{2}dx \bigg \}dx \\ \\ \longrightarrow \displaystyle \sf \dfrac{1}{3} {x}^{3} log(x) - \dfrac{1}{3} \int \dfrac{1}{x} \times {x}^{3} dx \\ \\ \longrightarrow \displaystyle \sf \dfrac{1}{3} {x}^{3} log(x) - \dfrac{1}{3} \int {x}^{2} dx \\ \\ \longrightarrow \displaystyle \sf \dfrac{1}{3} {x}^{3} log(x) - \dfrac{1}{3} \bigg \{ \dfrac{ {x}^{3} }{3} \bigg \} + c \\ \\ \longrightarrow \displaystyle \sf \dfrac{1}{3} {x}^{3} log(x) - \dfrac{ {x}^{3} }{9} + c$

Thus,

$\boxed{ \boxed{ \displaystyle \sf \int {x}^{2} log(x) dx = \dfrac{1}{3} {x}^{3} log(x) - \dfrac{ {x}^{3} }{9} + c }}$

Formula Used :

$\boxed{ \boxed{ \displaystyle \sf \int uvdx = u \int vdx - \int \bigg(u' \int vdx\bigg)dx }}$

Here, u and v are x dependant functions.