Indefinite Integrals — Question 2

Question 2

Find the most general antiderivative of $$\int \left( e^{2x}\sin x + \frac{1}{(1-x)^2} \right)\,dx.$$

Question 2 - Solution

We evaluate each term separately.

First term: $$\int e^{2x}\sin x\,dx.$$ Use integration by parts. Let $$u=\sin x, \qquad dv=e^{2x}\,dx.$$ Then $$du=\cos x\,dx, \qquad v=\frac{1}{2}e^{2x}.$$ So, $$\int e^{2x}\sin x\,dx = \frac{1}{2}e^{2x}\sin x - \frac{1}{2}\int e^{2x}\cos x\,dx.$$

Apply integration by parts again to the remaining integral. Let $$u=\cos x, \qquad dv=e^{2x}\,dx.$$ Then $$du=-\sin x\,dx, \qquad v=\frac{1}{2}e^{2x}.$$ Thus, $$\int e^{2x}\cos x\,dx = \frac{1}{2}e^{2x}\cos x + \frac{1}{2}\int e^{2x}\sin x\,dx.$$

Substitute this back: $$\int e^{2x}\sin x\,dx = \frac{1}{2}e^{2x}\sin x - \frac{1}{4}e^{2x}\cos x - \frac{1}{4}\int e^{2x}\sin x\,dx.$$

Add $\frac{1}{4}\int e^{2x}\sin x\,dx$ to both sides: $$\frac{5}{4}\int e^{2x}\sin x\,dx = \frac{1}{2}e^{2x}\sin x - \frac{1}{4}e^{2x}\cos x.$$

Solve: $$\int e^{2x}\sin x\,dx = \frac{e^{2x}}{5}(2\sin x - \cos x).$$

Second term: $$\int \frac{1}{(1-x)^2}\,dx.$$ Let $$u=1-x, \qquad du=-dx.$$ Then $$\int \frac{1}{(1-x)^2}\,dx = -\int u^{-2}\,du = u^{-1} = \frac{1}{1-x}.$$

Final Answer: $$\boxed{ \frac{e^{2x}}{5}(2\sin x - \cos x) + \frac{1}{1-x} + C }$$