Question

What is ${\int }_1^{e}x\ln xdx$ equal to?

• A. $\frac{e+1}{4}$
• B. $\frac{e^2+1}{4}$
• C. $\frac{e-1}{4}$
• D. $\frac{e^2-1}{4}$

Answer 1

Answer: (B)

$\frac{e^2+1}{4}$

Using Integration by parts:-

Consider 1st function $u\left(x\right)=lnx$ & 2nd function $v\left(x\right)=x$

Now using Integration by parts we get:-

${\int }_1^{e}x\ln xdx$ $=$ $\ln x{\int }_1^{e}xdx$ - ${\int }_1^e\left(\frac{d}{dx}\right(lnx\left){\int }_1^{e}xdx\right)$

$\Rightarrow$ $\left(\ln x\frac{x^2}{2}\right){|}_e^1$$-$${\int }_1^e(\frac{1}{x}\times \frac{x^2}{2})$

On putting limits and solving we get:-

$\Rightarrow$ $\frac{e^2}{2}-\frac{e^2}{4}+\frac{1}{4}$

$\Rightarrow \frac{e^2+1}{4}$