
Various exercises relating to integration by parts.

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.
(Add a constant to your answer if needed so that it equals $0$ at $x = 0$.)
(Add a constant to your answer if needed so that it equals $-1$ at $x = 0$.)
Don’t forget absolute values in your logarithm. (Add a constant to your answer as necessary so that it equals $-1/9$ at $x = 1$.)
(Add a constant to your answer if needed so that it equals $-1/2$ at $x = 0$.)
This is a case in which you need to treat the integration by parts as an equation and solve for the answer.
(Add a constant to your answer if needed so that it equals $1$ at $x = 0$.)
Write $\arcsin x = 1 \cdot \arcsin x$.

### Sample Quiz Questions

Compute the definite integral (Hints won’t be revealed until after you choose a response.)

$\displaystyle \frac {14}{9} e^{12} - \frac {5}{9} e^{3}$ $\displaystyle \frac {17}{9} e^{12} - \frac {5}{9} e^{3}$ $\displaystyle \frac {17}{9} e^{12} - \frac {8}{9} e^{3}$ $\displaystyle \frac {20}{9} e^{12} - \frac {8}{9} e^{3}$ $\displaystyle \frac {20}{9} e^{12} - \frac {11}{9} e^{3}$ $\displaystyle \frac {23}{9} e^{12} - \frac {11}{9} e^{3}$

Compute the definite integral (Hints won’t be revealed until after you choose a response.)

$\displaystyle 0$ $\displaystyle \frac {\pi }{4}$ $\displaystyle -\frac {\pi }{4}$ $\displaystyle \frac {3 \pi }{4}$ $\displaystyle -\frac {3 \pi }{4}$ $\displaystyle \frac {7 \pi }{4}$

Compute the indefinite integral (Hints won’t be revealed until after you choose a response.)

$\displaystyle x \arctan 5x - \frac {1}{10} \ln | 1 + 25x^2| + C$ $\displaystyle x \arctan 5x - \frac {1}{12} \ln | 1 + 25x^2| + C$ $\displaystyle x \arctan 5x - \frac {1}{14} \ln | 1 + 25x^2| + C$ $\displaystyle x \arctan 5x + \frac {1}{10} \ln | 1 + 25x^2| + C$ $\displaystyle x \arctan 5x + \frac {1}{12} \ln | 1 + 25x^2| + C$ $\displaystyle x \arctan 5x + \frac {1}{14} \ln | 1 + 25x^2| + C$

Compute the indefinite integral (Hints won’t be revealed until after you choose a response.)

$\displaystyle \frac {e^{3x} (2 \cos 5x + 5 \sin 5x)}{29} + C$ $\displaystyle \frac {e^{3x} (3 \cos 5x + 5 \sin 5x)}{34} + C$ $\displaystyle \frac {e^{3x} ( \cos 5x + 3 \sin 5x)}{20} + C$ $\displaystyle \frac {e^{3x} ( \cos 5x + 2 \sin 5x)}{15} + C$ $\displaystyle \frac {e^{3x} (2 \cos 5x + 7 \sin 5x)}{53} + C$ $\displaystyle \frac {e^{3x} (3 \cos 5x + 7 \sin 5x)}{58} + C$

### Sample Exam Questions

Compute the integral below.

$1 - \ln 2$ $2$ $\displaystyle \ln 2 - \frac {1}{2}$ $\displaystyle \frac {1}{2}$ $2 - 2 \ln 2$ the integral diverges

Compute the indefinite integral indicated below. [Hint: Write $\displaystyle \frac {1}{\cos ^2 \theta } = \sec ^2 \theta$ and integrate by parts.]

$\displaystyle (\sin \theta ) \ln |\sin \theta | + C$ $\displaystyle (\cos \theta ) \ln |\sin \theta | + C$ $\displaystyle (\tan \theta ) \ln |\sin \theta | + C$ $\displaystyle (\csc \theta ) \ln |\sin \theta | + C$ $\displaystyle (\sec \theta ) \ln |\sin \theta | + C$ $\displaystyle (\cot \theta ) \ln |\sin \theta | + C$