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} 😀


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 😀


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