Integration, CAS vs human

To evaluate integral

\int x(1+x)^{19}\; dx,

the CAS I have tried (maxima, mathematica) expands (1+x)^{19} first, followed by multiply the expression by x. Finally, integrate term by term yields the result, a messy looking polynomial (see Fig. 1)

Screen Shot 2017-05-17 at 8.39.07 PM.png

Fig. 1

A human being however, will more likely to recognize the fact that (1+x)^{19} is the derivative of \frac{(1+x)^{20}}{20} and therefore use the method of integration by parts:

\int x (1+x)^{19}\;dx

=\int x (\frac{(1+x)^{20}}{20})'\;dx

=x \frac{(1+x)^{20}}{20}-\int x'\frac{(1+x)^{20}}{20}\;dx

=x\frac{(1+x)^{20}}{20}-\frac{1}{20}\int(1+x)^{20}\;dx

=\frac{x(1+x)^{20}}{20}-\frac{(1+x)^{21}}{420}\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad (1)

An even better approach is:

\int x(1+x)^{19}\;dx

=\int (1+x-1) (1+x)^{19}\;dx

=\int ((1+x)-1) (1+x)^{19}\;dx

=\int (1+x)^{20}-(1+x)^{19}\;dx

=\int (1+x)^{20}\;dx-\int (1+x)^{19}\;dx

= \frac{(1+x)^{21}}{21}-\frac{(1+x)^{20}}{20}\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad(2)

The difference of result from CAS and (2) is a constant, as expected (see Fig. 2)

Screen Shot 2017-05-17 at 9.35.56 PM.png

Fig. 2

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s