$\text { Find the integral } \int \frac { t } { t ^ { 4 } + 81 } d t \text {. }$
A) $\frac { 1 } { 18 } \arctan 9 t ^ { 2 } d t + C$
B) $\frac { 1 } { 18 } \arctan \frac { t ^ { 2 } } { 9 } d t + C$
C) $\arctan \frac { t ^ { 2 } } { 81 } d t + C$
D) $\frac { 1 } { 9 } \arctan 81 t ^ { 2 } d t + C$
E) $\arctan \frac { t ^ { 2 } } { 9 } d t + C$

Question

$\text { Find the integral } \int \frac { t } { t ^ { 4 } + 81 } d t \text {. }$ 
A) $\frac { 1 } { 18 } \arctan 9 t ^ { 2 } d t + C$
B) $\frac { 1 } { 18 } \arctan \frac { t ^ { 2 } } { 9 } d t + C$
C) $\arctan \frac { t ^ { 2 } } { 81 } d t + C$
D) $\frac { 1 } { 9 } \arctan 81 t ^ { 2 } d t + C$
E) $\arctan \frac { t ^ { 2 } } { 9 } d t + C$

Quizplus · Accepted Answer

Use partial fraction decomposition:
$$\frac{t}{t^4+81}=\frac{1}{9}\left(\frac{t}{t^2+9}-\frac{t}{t^2-9}\right)$$
Then, use substitution $u=t^2-9$ for the second fraction, and $v=t^2+9$ for the first one:
$$\int \frac { t } { t ^ { 4 } + 81 } d t=\frac{1}{9}\left(\int \frac {1}{v} d v-\int \frac{1}{u+18} du \right) = \frac{1}{9}\ln|t^2+9|-\frac{1}{18}\ln|t^2-9|+C$$
And finally, by using the identity $\ln|a^2-b^2|=2\ln|a|-2\ln|b|$:
$$\int \frac { t } { t ^ { 4 } + 81 } d t=\frac{1}{9}\ln|t^2+9|-\frac{1}{9}\ln|t+3|+\frac{1}{9}\ln|t-3|+C=\boxed{\frac { 1 } { 18 } \arctan \frac { t ^ { 2 } } { 9 } }+C$$

Find the integral ∫tt4+81dt. \text { Find the integral } \int \frac { t } { t ^ { 4 } + 81 } d t \text {. } Find the integral ∫t4+81t​dt.

$\text { Find the integral } \int \frac { t } { t ^ { 4 } + 81 } d t \text {. }$