OBJECTIVES

When you finish this section, you should be able to:

1. Use a Table of Integrals (p. 520)
2. Use a computer algebra system (p. 522)

Use a Table of Integrals to find $\int \dfrac{dx}{\sqrt{\left(4x-x^{2}\right) ^{3}}}.$

Solution Look through the headings in the Table of Integrals until you locate Integrals Containing $\sqrt{2ax-x^{2}}.$ Continue in the subsection until you find a form that closely resembles the integrand given. You should find Integral 82: $$\int \dfrac{dx}{(2ax-x^{2}) ^{3/2}}=\dfrac{x-a}{a^{2}\sqrt{2ax-x^{2}}}+C$$ This is the integral we seek with $a = 2.$ So, $$\int \dfrac{dx}{\sqrt{( 4x-x^{2}) ^{3}}}=\dfrac{x-2}{4\sqrt{4x-x^{2}}}+C$$

NOW WORK

Problem 3.

Use a Table of Integrals to find $\int x^{2}\tan ^{-1}x\,dx$.

Solution Find the subsection of the table titled Integrals Containing Inverse Trigonometric Functions. Then look for an integral whose form closely resembles the problem. You should find Integral 114: $$\int x^{n}\,\tan ^{-1}x\,dx=\dfrac{1}{n+1} \left( x^{n+1}\,\tan ^{-1}x-\int \dfrac{x^{n+1}\,dx}{1+x^{2}}\right)\qquad n\neq -1$$

This is the integral we seek with $n=2.$ $$\int x^{2} \tan ^{-1}x~dx = \dfrac{1}{3}\left( x^{3} \tan^{-1}x-\int \dfrac{x^{3} dx}{1+x^{2}}\right)$$

We find the integral on the right by using the substitution $u=1+x^{2}.$ Then $du=2x\,dx$ and $$\begin{eqnarray*} \int \dfrac{x^{3}}{1+x^{2}}dx &=& \int \dfrac{x^{2}x\,dx}{1+x^{2}}=\int \dfrac{u-1 }{u}\dfrac{du}{2}=\dfrac{1}{2}\int \left( 1-\dfrac{1}{u}\right) du=\dfrac{1}{ 2}u-\dfrac{1}{2}\ln \vert u \vert \\[6pt] &=& \dfrac{1+x^{2}}{2}-\dfrac{\ln (1+x^{2}) }{2} \end{eqnarray*}$$

So, $$\begin{eqnarray*} \int x^{2}\,\tan ^{-1}x\,dx &=& \dfrac{1}{3}\left( x^{3}\,\tan ^{-1}x-\int \dfrac{x^{3}}{1+x^{2}}\,dx\right) \\[6pt] &=& \dfrac{1}{3}\left[ x^{3}\,\tan ^{-1}x- \dfrac{1+x^{2}}{2}+\dfrac{1}{2}\ln ( 1+x^{2}) \right] +C \end{eqnarray*}$$

NOW WORK

Problem 11.

Use a Table of Integrals to find $\int \dfrac{3x+5}{\sqrt{3x+6}}\,dx$.

Solution Find the subsection of the table titled Integrals Containing $\sqrt{ax+b}$ (the square root of a linear expression). Then look for an integral whose form closely resembles the problem. The closest one is an integral with $x$ in the numerator and $\sqrt{ax+b}$ in the denominator, $$\begin{equation*} \hbox{Integral 42:}\qquad\int \dfrac{x~dx}{\sqrt{a+bx}}=\dfrac{2}{3b^{2}}(bx-2a) \sqrt{a+bx}+C\tag{1} \end{equation*}$$

To express the given integral as one with a single variable in the numerator, use the substitution $u=3x+5.$ Then $du=3\,{\it dx}.$ Since $$\sqrt{3x+6}=\sqrt{(3x+5) +1}=\sqrt{u+1}$$

we find $$\int \dfrac{3x+5}{\sqrt{3x+6}}\,dx \underset{\underset{\color{#0066A7}{\hbox{\(u=3x+5, \tfrac{1}{3}du=dx\)}}}{\color{#0066A7}{\uparrow }}}{=} \dfrac{1}{3}\int \dfrac{u\,du}{\sqrt{u+1}} \\ $$

which is in the form of (1) with $a=1$ and $b=1$ So, $$\begin{align*} \int \dfrac{3x+5}{\sqrt{3x+6}}\,dx = \dfrac{1}{3}\int \dfrac{udu}{\sqrt{u+1}} \underset{\underset{\color{#0066A7}{\hbox{(1)}}}{{\color{#0066A7}\uparrow}}}{=}\dfrac{1}{3} \cdot \dfrac{2}{3} (u-2) \sqrt{1+u}+C \\ \underset{\underset{\color{#0066A7}{u=3x+5}}{\color{#0066A7}{\uparrow}}}{=}\dfrac{2}{9}[(3x+5) -2] \sqrt{1+(3x+5)}+C=\dfrac{2}{3}(x+1) \sqrt{3x+6}+C \end{align*}$$

NOW WORK

Problem 5.

NEED TO REVIEW?

Computer algebra systems are discussed in Section P.8, pp. 64-67.

Find $\int x^{2}( 2x^{3}-4) ^{5}dx$.

Solution Using WolframAlpha to find the integral, input $$\hbox{integrate }x^{\wedge }2((2x^{\wedge }3 ) -4)^{\wedge }5$$

The output is $$\int x^{2}( 2x^{3}-4) ^{5}dx=32\left( \dfrac{x^{18}}{18}-\dfrac{ 2x^{15}}{3}+\dfrac{10x^{12}}{3}-\dfrac{80x^{9}}{9}+\dfrac{40x^{6}}{3}-\dfrac{ 32x^{3}}{3}\right) +C$$

Using Mathematica returns the same result without the constant $C$.

NOW WORK

Problems 19, 21, and 27 using a CAS and compare your answers to Problems 3, 5, and 11.