Integration by parts
Example 1:
$$\int xe^x \, dx $$
Solution:
$$\int xe^x \, dx $$
\begin{align*}
\textup{Let}\,\, u &= x \;\;\;\;\;\;\;\; dv = e^x \,dx\\
du &= dx \;\;\;\;\;\;\;\; v = e^x \\\\
\because\int u\,dv &= uv-\int v\,du \\\\
\therefore\int xe^x \,dx &= xe^x-\int e^x\,dx \\
\int xe^x \,dx &= xe^x-e^x+C \\
\end{align*}
Example 2:
$$\int x \sin x \, dx $$
Solution:
$$\int x \sin x \, dx $$
\begin{align*}
\textup{Let}\,\, u &= x \;\;\;\;\;\;\;\; dv = \sin x \,dx\\
du &= dx \;\;\;\;\;\;\;\; v = -\cos x \\\\
\because\int u\,dv &= uv-\int v\,du \\\\
\therefore\int x \sin x \,dx &= x (-\cos x) -\int (-\cos x)\,dx \\
\int x \sin x \,dx &= -x \cos x + \int \cos x\,dx \\
\int x \sin x \,dx &= -x \cos x+ \sin x+C \\
\end{align*}
Example 3:
$$\int x \cos x \, dx $$
Solution:
$$\int x \cos x \, dx $$
\begin{align*}
\textup{Let}\,\, u &= x \;\;\;\;\;\;\;\; dv = \cos x \,dx\\
du &= dx \;\;\;\;\;\;\;\; v = \sin x \\\\
\because\int u\,dv &= uv-\int v\,du \\\\
\therefore\int x \cos x \,dx &= x \sin x -\int \sin x\,dx \\
\int x \cos x \,dx &= x \sin x - (-\cos x) + C \\
\int x \cos x \,dx &= x \sin x + \cos x + C \\
\end{align*}
Example 4:
$$\int xe^{-x} \, dx $$
Solution:
$$\int xe^{-x} \, dx $$
\begin{align*}
\textup{Let}\,\, u &= x \;\;\;\;\;\;\;\; dv = e^{-x} \,dx\\
du &= dx \;\;\;\;\;\;\;\; v = -e^{-x} \\\\
\because\int u\,dv &= uv-\int v\,du \\\\
\therefore\int xe^{-x} \,dx &= x(-e^{-x})-\int (-e^{-x}) \,dx \\
\int xe^{-x} \,dx &= -xe^{-x} + \int e^{-x} \,dx \\
\int xe^{-x} \,dx &= -xe^{-x} - e^{-x} + C \\
\end{align*}
Example 5:
$$\int \ln x \, dx $$
Solution:
$$\int \ln x \, dx $$
\begin{align*}
\textup{Let}\,\, u &= \ln x \;\;\;\;\;\;\;\; dv = dx\\
du &= \frac{1}{x} dx \;\;\;\;\;\;\;\; v = x\\\\
\because\int u\,dv &= uv-\int v\,du \\\\
\therefore\int \ln x \,dx &= (\ln x)(x) -\int (x)\left(\frac{1}{x} \,dx \right) \\
\int \ln x \,dx &= x \ln x- \int 1 \,dx \\
\int \ln x \,dx &= x \ln x - x + C \\
\end{align*}
Example 7:
$$\int \ln x^2 \, dx $$
Solution:
\begin{align*}
\textup{Let } I &=\int \ln x^2 \, dx \\
I&= \int 2\ln x \, dx \\
I&= 2 \int \ln x \, dx \\\\
\textup{Consider } &\int \ln x \, dx\\\\
\textup{Let }\,\, u &= \ln x \;\;\;\;\;\;\;\; dv = dx\\
du &= \frac{1}{x} dx \;\;\;\;\;\;\;\; v = x\\\\
\because\int u\,dv &= uv-\int v\,du \\\\
\therefore\int \ln x \,dx &= (\ln x)(x) -\int (x)\left(\frac{1}{x} \,dx \right) \\
\int \ln x \,dx &= x \ln x- \int 1 \,dx \\
\int \ln x \,dx &= x \ln x - x + K \\\\
\therefore I&= 2 \int \ln x \, dx \\
\int \ln x^2 \, dx &= 2(x \ln x - x + K)\\
&= 2x \ln x - 2x + 2K\\
&= x\times 2 \ln x - 2x + C \;\;(\textup{where }C=2K) \\
&= x \ln x^2 - 2x + C\\
\end{align*}