Evaluate the integral by making a substitution and then using a table of integrals.
-$$\int \frac { d x } { x \left( 36 + ( \ln x ) ^ { 2 } \right) }$$
A) $\frac { 1 } { 6 } \tan ^ { - 1 } \left( \frac { \ln x } { 6 } \right) + C$
B) $\frac { 1 } { 6 } \sin ^ { - 1 } \left( \frac { x } { 6 } \right) + C$
C) $\frac { 1 } { 6 } \tan ^ { - 1 } \left( \frac { \mathrm { x } } { 6 } \right) + \mathrm { C }$
D) $\frac { 1 } { 6 } \sin ^ { - 1 } \left( \frac { \ln x } { 6 } \right) + C$

Question

Quizplus · Accepted Answer

To solve the integral, make the substitution $u = \ln(x)$, which implies $du = \frac{dx}{x}$. The integral becomes $\int \frac{du}{36+u^2}$, which matches the form of the integral that leads to the inverse tangent function, resulting in $\frac{1}{6}\tan^{-1}\left(\frac{u}{6}\right) + C$, or substituting back for $u$, $\frac{1}{6}\tan^{-1}\left(\frac{\ln(x)}{6}\right) + C$.

Evaluate the Integral by Making a Substitution and Then Using ∫dxx(36+(ln⁡x)2)\int \frac { d x } { x \left( 36 + ( \ln x ) ^ { 2 } \right) }∫x(36+(lnx)2)dx​

Evaluate the Integral by Making a Substitution and Then Using $\int \frac { d x } { x \left( 36 + ( \ln x ) ^ { 2 } \right) }$