Question

${\int }_1^2{x}^{2x}[1+\log x]dx=$

• A. $\frac{15}{2}$
• B. $\frac{13}{2}$
• C. $\frac{11}{2}$
• D. $\frac{9}{2}$

Answer 1

Answer: (A)

Let $x^x=t$
Diiferentiating both sides,
$x^x(1+\log x)dx=dt$
Substituting the above obtained values back into the expression,
$\int x^{2x}(1+logx)dx=\int tdt$
$=\frac{t^2}{2}+c$
Putting the value of t back, we get the integral as
$\frac{{\left(x^x\right)}^2}{2}+c$
Substituting the limits from 1 to 2, we get
$=\left(\frac{{\left(2^2\right)}^2}{2}\right)-\left(\frac{{\left(1^1\right)}^2}{2}\right)$
.
$=8-\frac{1}{2}$
.
$=\frac{15}{2}$