## 12th Harvard MIT Calculus Problem 8

Compute 😀 $A=\int\limits_{1}^{\sqrt{3}}{{{x}^{2{{x}^{2}}+1}}+\ln ({{x}^{2{{x}^{2{{x}^{2}}+1}}}})dx}$ 😀

Solution

Using the fact that $x={{e}^{\ln x}}$, we evaluate the integral as follows:

$\int{{{x}^{2{{x}^{2}}+1}}+\ln ({{x}^{2{{x}^{2{{x}^{2}}+1}}}})dx=\int{{{x}^{2{{x}^{2}}+1}}+{{x}^{2{{x}^{2}}+1}}(\ln {{x}^{2}})dx}}$

$A=\int{{{e}^{\ln (x)(2{{x}^{2}}+1)}}(1+\ln {{x}^{2}})dx}$

$A=\int{x{{e}^{{{x}^{2}}\ln {{x}^{2}}}}(1+\ln {{x}^{2}})dx}$

Noticing that the derivative of ${{x}^{2}}\ln {{x}^{2}}$ is $2x\ln (1+{{x}^{2}})$, it follows that the integral evaluates to $\frac{1}{2}{{e}^{{{x}^{2}}\ln {{x}^{2}}}}=\frac{1}{2}{{x}^{2{{x}^{2}}}}$

Evaluate this from $1$ to $\sqrt{3}$ we obtain the answer $\int\limits_{1}^{\sqrt{3}}{{{x}^{2{{x}^{2}}+1}}+\ln ({{x}^{2{{x}^{2{{x}^{2}}+1}}}})dx}=\int\limits_{0}^{3\ln 3}{{{e}^{u}}du=13}$ 😀

Remember to change from x to u 😀