Evaluate the integral.
-$$\int \frac { ( x + 7 ) ^ { 2 } \tan ^ { - 1 } x + ( 4 x - 22 ) ( x + 7 ) } { \left( x ^ { 2 } + 1 \right) ( x + 7 ) ^ { 2 } } d x$$
A) $\frac { \left( \tan ^ { - 1 } x \right) ^ { 2 } } { 2 } - \ln | x + 7 | + \frac { 1 } { 2 } \ln \left( x ^ { 2 } + 1 \right) + C$
B) $\frac { \left( \tan ^ { - 1 } x \right) ^ { 2 } } { 2 } - \ln | x + 7 | - 5 \tan ^ { - 1 } x + \ln \left( x ^ { 2 } + 1 \right) + C$
C) $\frac { \left( \tan ^ { - 1 } x \right) ^ { 2 } } { 2 } - 3 \tan ^ { - 1 } x + \frac { 1 } { 2 } \ln \left( x ^ { 2 } + 1 \right) + C$
D) $\frac { \left( \tan ^ { - 1 } x \right) ^ { 2 } } { 2 } - \ln | x + 7 | - 3 \tan ^ { - 1 } x + \frac { 1 } { 2 } \ln \left( x ^ { 2 } + 1 \right) + C$

Question

Quizplus · Accepted Answer

The integral can be solved by breaking it into parts and using substitution. The first part involves $(x+7)^2 \tan^{-1}x$ which integrates to $\frac{(\tan^{-1}x)^2}{2}$. The second part involves $(4x-22)(x+7)$ which simplifies to $-\ln|x+7| - 3\tan^{-1}x + \frac{1}{2}\ln(x^2+1)$.

Evaluate the Integral ∫(x+7)2tan⁡−1x+(4x−22)(x+7)(x2+1)(x+7)2dx\int \frac { ( x + 7 ) ^ { 2 } \tan ^ { - 1 } x + ( 4 x - 22 ) ( x + 7 ) } { \left( x ^ { 2 } + 1 \right) ( x + 7 ) ^ { 2 } } d x∫(x2+1)(x+7)2(x+7)2tan−1x+(4x−22)(x+7)​dx

Evaluate the Integral $\int \frac { ( x + 7 ) ^ { 2 } \tan ^ { - 1 } x + ( 4 x - 22 ) ( x + 7 ) } { \left( x ^ { 2 } + 1 \right) ( x + 7 ) ^ { 2 } } d x$