§5.5 Other options for finding algebraic antiderivatives

Introduction

The Method of Partial Fractions

Using an Integral Table

Using Computer Algebra Systems

Summary

Given $\frac{1}{x^2-2x-3} = \frac{1/4}{x-3} - \frac{1/4}{x+1}$, the integral $\int\frac{1}{x^2-2x-3}\,dx$ equals

The integral $\int\frac{x}{\sqrt{x^2+4}}\,dx$ is best evaluated by

Among $\int x^2\sin(x^3)\,dx$, $\int x^2\sin(x)\,dx$, $\int\sin(x^3)\,dx$, and $\int x^5\sin(x^3)\,dx$, the correct classifications are

Using the decomposition $\frac{5x}{x^2-x-2} = \frac{10/3}{x-2} + \frac{5/3}{x+1}$, the integral $\int\frac{5x}{x^2-x-2}\,dx$ equals

Given $\frac{x^2+1}{x^3-x^2} = -\frac{1}{x} - \frac{1}{x^2} + \frac{2}{x-1}$, the integral $\int\frac{x^2+1}{x^3-x^2}\,dx$ equals

Given $\frac{x-2}{x^4+x^2} = \frac{1}{x} - \frac{2}{x^2} + \frac{-x+2}{1+x^2}$, the integral $\int\frac{x-2}{x^4+x^2}\,dx$ equals

Using the integral table with $a = 2$, the antiderivative $\int\sqrt{x^2+4}\,dx$ equals

Using $a = 4$ and $u = 5x$, the integral $\int\frac{2}{\sqrt{16+25x^2}}\,dx$ equals

With $a = 7$ and $u = 6x$, the table shows $\int\frac{1}{x^2\sqrt{49-36x^2}}\,dx$ equals

To evaluate $\int\sqrt{9+64x^2}\,dx$ with the table, the right substitution is

The decomposition $\frac{x^2-x-1}{(x-3)^3} = \frac{1}{x-3} + \frac{5}{(x-3)^2} + \frac{5}{(x-3)^3}$ yields the integral

For $\int x\sqrt{1+x^4}\,dx$, the opening move is

For the integral $\int\frac{\sqrt{x+\sqrt{1+x^2}}}{x}\,dx$, trying $u = 1+x^2$, $u = x+\sqrt{1+x^2}$, and by parts all fail because

Using $\int\frac{x}{\sqrt{x^2+4}}\,dx = \sqrt{x^2+4} + C$, evaluate $\int_0^1\frac{x}{\sqrt{x^2+4}}\,dx$ as a decimal, rounded to three places.

Using $\int\frac{1}{x^2-2x-3}\,dx = \frac{1}{4}\ln|x-3| - \frac{1}{4}\ln|x+1| + C$, evaluate $\int_4^5\frac{1}{x^2-2x-3}\,dx$, rounded to three decimals. The exact value is $\frac{1}{4}\ln\frac{5}{3}$.

When decomposing a rational function whose denominator contains a repeated linear factor like $x^2$, the decomposition needs separate $\frac{A}{x}$ and $\frac{B}{x^2}$ terms.