Integration by Substitution
Integration by Substitution
Example 1:
$$\int (x^2+1)^4 \, xdx $$
Solution:
$$\int (x^2+1)^4 \, xdx $$
\begin{align*}
\textup{Let}\,\, u & =x^2+1 \\
du & =2xdx \\
\frac{du}2 & =xdx \\
\int (x^2+1)^4 \, xdx &= \int u^4\,\, \frac{du}2 \\
& = \frac{1}2 \int u^4\,du \\
& = \frac{1}2 \, ( \frac{u^{4+1}}{4+1})+C \\
& = \frac{1}2 \, ( \frac{u^{5}}{5})+C \\
& = \frac{1}{10} \, u^5+C \\
& = \frac{1}{10} \, (x^2+1)^5+C \\
\end{align*}
Example 2:
$$\int (x^3+1)^5 \, x^2dx $$
Solution:
$$\int (x^3+1)^5 \, x^2dx $$
\begin{align*}
\textup{Let}\,\, u & =x^3+1\\
du & =3x^2dx\\
\frac{du}3 & = x^2 dx \\
\int (x^3+1)^5 \, x^2 dx &= \int u^5\,\, \frac{du}3 \\
& = \frac{1}3 \int u^5\,du \\
& = \frac{1}3 \, \left( \frac{u^{5+1}}{5+1}\right)+C \\
& = \frac{1}3 \, \left( \frac{u^{6}}{6}\right)+C \\
& = \frac{1}{18} \, u^6+C \\
& = \frac{1}{18} \, (x^3+1)^6+C \\
\end{align*}
Example 3:
$$\int \frac{x}{\sqrt{x^2+1}}\,dx $$
Solution:
\begin{align*}
\int \frac{x}{\sqrt{x^2+1}}\,dx &=\int \frac{1}{\sqrt{x^2+1}}x\,dx \\\\
\textup{Let}\,\, u & =x^2+1\\
du & =2xdx\\
\frac{du}2 & = x dx \\\\
\int \frac{x}{\sqrt{x^2+1}}\,dx &=\int \frac{1}{\sqrt{u}}\,\frac{du}{2}\\
&=\frac{1}{2}\int \frac{1}{u^{\frac{1}{2}}}\,du\\
&=\frac{1}{2}\int u^{-\frac{1}{2}}\,du\\
& = \frac{1}{2}\, \left( \frac{u^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\right)+C \\
& = \frac{1}{2}\, \left( \frac{u^{\frac{1}{2}}}{\frac{1}{2}}\right)+C \\
& = u^{\frac{1}{2}}+C \\
& = \sqrt{u}+C \\
& = \sqrt{x^2+1}+C \\
\end{align*}