Suggested languages for you:

Americas

Europe

Q. 78

Expert-verifiedFound in: Page 465

Book edition
1st

Author(s)
Peter Kohn, Laura Taalman

Pages
1155 pages

ISBN
9781429241861

Solve given integrals by using polynomial long division to rewrite the integrand. This is one way that you can sometimes avoid using trigonometric substitution; moreover, sometimes it works when trigonometric substitution does not apply.

$\int \frac{{x}^{3}-3{x}^{2}+2x-3}{{x}^{2}+1}$dx

$\frac{{x}^{2}}{2}-3x+\frac{1}{2}\mathrm{ln}\left(\left|{x}^{2}+1\right|\right)+C\text{.}$

The integral is as follows.

$\int \frac{{x}^{3}-3{x}^{2}+2x-3}{{x}^{2}+1}dx$

The objective is ton solve the integral.

The polynomial long division method is calculated below.

$x-3\phantom{\rule{0ex}{0ex}}\begin{array}{c}{x}^{2}+1\overline{){x}^{3}-3{x}^{2}+2x-3}\\ \frac{{x}^{3}+x}{-3{x}^{2}+x}\\ \frac{-3{x}^{2}-3}{x}\end{array}$

The expression is in the form of $\frac{{x}^{3}-3{x}^{2}+2x-3}{{x}^{2}+1}=x-3+\frac{x}{{x}^{2}+1}$

The integral is solved below.

$\begin{array}{l}\int \frac{{x}^{3}-3{x}^{2}+2x-3}{{x}^{2}+1}dx\\ =\int \left(x-3+\frac{x}{{x}^{2}+1}\right)dx\\ =\int xdx-\int 3dx+\int \frac{x}{{x}^{2}+1}dx\\ =\frac{{x}^{2}}{2}-3x+\frac{1}{2}\int \frac{1}{u}du\phantom{\rule{1em}{0ex}}\left[u={x}^{2}+1,du=2xdx\right]\\ =\frac{{x}^{2}}{2}-3x+\frac{1}{2}\mathrm{ln}\left(\right|u\left|\right)+C\phantom{\rule{1em}{0ex}}\frac{{x}^{2}}{2}-3x+\frac{1}{2}\mathrm{ln}\left(\left|{x}^{2}+1\right|\right)+C\end{array}$

Therefore, the value is $\frac{{x}^{2}}{2}-3x+\frac{1}{2}\mathrm{ln}\left(\left|{x}^{2}+1\right|\right)+C\text{.}$

94% of StudySmarter users get better grades.

Sign up for free