Exercises: Substitution and Tables

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Various exercises relating to substitution and the use of integral tables.

Compute the indefinite integrals below. Since there are many possible answers (which differ by constant values), use the given instructions if needed to choose which possible answer to use.

$\int (12 x+14) \left (3 x^2+7 x-1\right )^5 dx = \answer {\frac {1}{3}(3 x^2+7 x-1)^6}+C$ (Add a constant to your answer if needed so that it equals $1/3$ at $x = 0$ .)

$\int \frac {e^{\sqrt {x}}}{\sqrt {x}} dx = \answer {2 e^{\sqrt {x}}}+C$ (Add a constant to your answer if needed so that it equals $2$ at $x = 0$ .)

$\int \frac {x}{\sqrt {x+3}} dx = \answer {\frac {2}{3} (x-6) \sqrt {x+3}}+C$ (Add a constant to your answer if needed so that it equals $0$ at $x=6$ .)

$\int \frac {\ln |x|}{x} dx = \answer {\frac {1}{2}\ln ^2 |x|}+C$ Remember absolute value in your logarithm. (Add a constant to your answer if needed so that it equals $0$ at $x = 1$ .)

$\int \sin x \sqrt {\cos x} dx = \answer {-\frac {2}{3} \cos ^{\frac {3}{2}}(x)}+C$ (Add a constant to your answer if needed so that it equals $-2/3$ at $x = 0$ .)

$\int \frac {9 (2 x+3)}{3 x^2+9 x+7} dx = \answer {3 \ln \left |3 x^2+9 x+7\right |}+C$ Use absolute values as needed in logarithms. (Add a constant to your answer if needed so that it equals $3 \ln 7$ at $x = 0$ .)

$\int \frac {x}{x^4+81} dx = \answer {\frac {1}{18} \arctan \left (\frac {x^2}{9}\right )}+C$ (Add a constant to your answer if needed so that it equals $0$ at $x = 0$ .)

Make a substitution $x^2 = 9 u$ .

Evaluate the definite integral $\displaystyle \int _{-2}^{-1} (x+1)e^{x^2+2x+1}\ dx.$

Value = $\answer {(1-e)/2}$

Sample Exam Questions

Evaluate the integral $\displaystyle \int _1^3 \left ( x - \sqrt {4 x^2 - 8 x + 13} \right ) dx$ using the fact that $\displaystyle \int _0^4 \sqrt {x^2 + 9} ~ dx = \frac {20 + 9 \ln 3}{2}$ . (Hints will not be displayed until you have chosen a response.)

$\displaystyle -\frac {4 + 9 \ln 3}{6}$ $\displaystyle -\frac {4 + 9 \ln 3}{4}$ $\displaystyle -\frac {4 + 9 \ln 3}{2}$ $\displaystyle -4 - 9 \ln 3$ $\displaystyle -8 - 18 \ln 3$ $\displaystyle -12 - 27 \ln 3$

First complete the square: $4 x^2 - 8 x + 13 = 4 x^2 - 8 x + 4 + (13 - 4) = (2x - 2)^2 + 9.$

Next substitute $u = 2x-2$ (i.e., $x = \frac {u}{2} + 1$ ). Then $du = 2 dx$ and $x=1 \leftrightarrow u = 0$ , $x=3 \leftrightarrow u = 4$ , so $\int _1^3 \left ( x - \sqrt {4x^2 - 8x + 13} \right ) dx = \int _0^4 \left ( \frac {u}{2} + 1 - \sqrt {u^2 + 9} \right ) \frac {du}{2}.$

Now use linearity of the integral to finish: $\int _0^4 \frac {u}{4} ~ du + \int _0^4 \frac {1}{2} ~ du - \int _0^4 \frac {1}{2} \sqrt {u^2+9} ~ du = \left . \frac {u^2}{8} \right |_0^4 + \left . \frac {u}{2} \right |_0^4 - \frac {20 + 9 \ln 3}{4} = 4 - \frac {20+9 \ln 3}{4} = - \frac {4 + 9 \ln 3}{4}.$