**Exercise:** Evaluate the indefinite integral.

- \int x\sin(x^2)dx
- \int e^x\sqrt{1+e^x}dx
- \int \sqrt{\cot x}\csc^2xdx
- \int \frac{dx}{\sqrt{1-x^2}\sin^{-1}x}dx
- \int \frac{\sin 2x}{1+cos^2x}dx
- \int \frac{1+x}{1+x^2}dx

- \int x\sin(x^2)dx=\int \frac{1}{2}\sin(x^2)dx^2=\frac{-1}{2}\cos(x^2)+C
- \int e^x\sqrt{1+e^x}dx=\int \sqrt{1+e^x}d(1+e^x)=\frac{2}{3}(1+e^x)^{3/2}+C
- \int \sqrt{\cot x}\csc^2xdx=\int -\sqrt{\cot x}d \cot x=\frac{-2}{3}(\cot x)^{3/2}+C
- \int \frac{dx}{\sqrt{1-x^2}\sin^{-1}x}dx=\int \frac{d\sin^{-1}x}{\sin^{-1}x}=ln|\sin^{-1}x|+C
- \int \frac{\sin 2x}{1+cos^2x}dx=\int \frac{2\sin x\cos x}{1+\cos^2x}dx=\int \frac{-d(1+cos^2x)}{1+\cos^2x}=\ln(1+\cos^2x)+C
- \int \frac{1+x}{1+x^2}dx=\int \frac{1}{1+x^2}dx+\int\frac{1}{2}\frac{d(1+x^2)}{1+x^2}dx=\tan^{-1}x+\frac{1}{2}ln(1+x^2)+C

**Exercise:** Evaluate the difinite integral. \int_1^2\frac{e^{1/x}}{x^2}dx.

Let u=1/x. By the substitution rule \int_1^2 \frac{e^{1/x}}{x^2}dx=\int_1^{1/2}-e^{u}du=\int_{1/2}^1 e^u du=e-e^{1/2}.