Find the indefinite integral. $$\int \frac { 1 } { 36 + ( x - 4 ) ^ { 2 } } d x$$
A) $\frac { 1 } { 36 } \arctan \left( \frac { x - 4 } { 36 } \right) + C$
B) $36 \arctan \left( \frac { x - 4 } { 36 } \right) + C$
C) $\frac { 1 } { 6 } \arctan \left( \frac { x - 4 } { 6 } \right) + C$
D) $\frac { 1 } { 36 } \arctan \left( \frac { x - 4 } { 6 } \right) + C$
E) $6 \arctan \left( \frac { x - 4 } { 6 } \right) + C$

Question

Quizplus · Accepted Answer

The integral is of the form $\int \frac{1}{a^2 + u^2} du$, which integrates to $\frac{1}{a} \arctan\left(\frac{u}{a}\right) + C$. Here, $a^2 = 36$, so $a = 6$, and $u = x - 4$, leading to the correct answer $\frac{1}{6} \arctan\left(\frac{x - 4}{6}\right) + C$.

Find the Indefinite Integral ∫136+(x−4)2dx\int \frac { 1 } { 36 + ( x - 4 ) ^ { 2 } } d x∫36+(x−4)21​dx

Find the Indefinite Integral $\int \frac { 1 } { 36 + ( x - 4 ) ^ { 2 } } d x$