Integration by Parts — Question 1

Question 1

Evaluate by tabular integration: $$\int x^3e^{2x}\,dx.$$

Question 1 – Solution

We use tabular integration because repeatedly differentiating the polynomial $x^3$ eventually gives zero. Differentiate in the $D$ column and integrate in the $I$ column. The signs alternate, beginning with a plus sign.

$$\begin{array}{c|c|c} \text{Sign} & D & I \\ \hline + & x^3 & \dfrac{e^{2x}}{2} \\[4pt] - & 3x^2 & \dfrac{e^{2x}}{4} \\[4pt] + & 6x & \dfrac{e^{2x}}{8} \\[4pt] - & 6 & \dfrac{e^{2x}}{16} \\[4pt] & 0 & \end{array}$$

For example, each entry in the integration column is obtained using $$\int e^{2x}\,dx=\frac{1}{2}e^{2x},$$ so every step down that column introduces another factor of $\frac12$.

Now multiply across each row, pairing each entry in the derivative column with the entry in the integration column on the same row, and use the indicated signs: $$\begin{align*} \int x^3e^{2x}\,dx &=x^3\left(\frac{e^{2x}}{2}\right) -3x^2\left(\frac{e^{2x}}{4}\right) +6x\left(\frac{e^{2x}}{8}\right) -6\left(\frac{e^{2x}}{16}\right)+C \\[4pt] &=\frac{x^3e^{2x}}{2} -\frac{3x^2e^{2x}}{4} +\frac{6xe^{2x}}{8} -\frac{6e^{2x}}{16}+C \\[4pt] &=e^{2x}\left(\frac{x^3}{2}-\frac{3x^2}{4} +\frac{3x}{4}-\frac{3}{8}\right)+C. \end{align*}$$

Therefore, $$\boxed{\displaystyle \int x^3e^{2x}\,dx =e^{2x}\left(\frac{x^3}{2}-\frac{3x^2}{4} +\frac{3x}{4}-\frac38\right)+C}$$

As a check, differentiating the answer gives $$\frac{d}{dx}\left[ e^{2x}\left(\frac{x^3}{2}-\frac{3x^2}{4} +\frac{3x}{4}-\frac38\right)\right] =x^3e^{2x},$$ which is the original integrand.