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*}