Indefinite Integrals — Question 1

Question 1

Evaluate the indefinite integral $$\int \left( x^2 e^{x^3} + \frac{1}{x}\sqrt{1+\ln x} - \frac{3x^2}{1+x^3} \right)\,dx.$$

Question 1 - Solution

We integrate term by term.

First term: $$\int x^2 e^{x^3}\,dx.$$ Let $u=x^3$, so $du=3x^2\,dx$: $$\int x^2 e^{x^3}\,dx = \frac{1}{3}\int e^{u}\,du = \frac{1}{3}e^{x^3}.$$

Second term: $$\int \frac{1}{x}\sqrt{1+\ln x}\,dx.$$ Let $u=1+\ln x$, so $du=\frac{1}{x}\,dx$: $$\int \sqrt{u}\,du = \frac{2}{3}u^{3/2} = \frac{2}{3}(1+\ln x)^{3/2}.$$

Third term: $$\int \frac{3x^2}{1+x^3}\,dx.$$ Let $u=1+x^3$, so $du=3x^2\,dx$: $$\int \frac{3x^2}{1+x^3}\,dx = \int \frac{du}{u} = \ln(1+x^3).$$

Final Answer: $$\boxed{ \frac{1}{3}e^{x^3} + \frac{2}{3}(1+\ln x)^{3/2} - \ln(1+x^3) + C }$$