The integral $\int \frac { 1 } { \sqrt { 16 x ^ { 2 } + 1 } } d x$ is

A) $\frac { \sinh ^ { - 1 } ( 4 x ) } { - 4 } + C$
B) $\frac { \sinh ^ { - 1 } ( 4 x ) } { 2 } + C$
C) $\frac { \sinh ^ { - 1 } ( 4 x ) } { 4 } + C$
D) $\frac { \sinh ^ { - 1 } ( 4 x ) } { - 2 } + C$
E) $\frac { \sinh ^ { - 1 } ( 4 x ) } { 8 } + C$

Question

The integral $\int \frac { 1 } { \sqrt { 16 x ^ { 2 } + 1 } } d x$ is

A) $\frac { \sinh ^ { - 1 } ( 4 x ) } { - 4 } + C$
B) $\frac { \sinh ^ { - 1 } ( 4 x ) } { 2 } + C$
C) $\frac { \sinh ^ { - 1 } ( 4 x ) } { 4 } + C$
D) $\frac { \sinh ^ { - 1 } ( 4 x ) } { - 2 } + C$
E) $\frac { \sinh ^ { - 1 } ( 4 x ) } { 8 } + C$

Quizplus · Accepted Answer

The integral $\int \frac { 1 } { \sqrt { 16 x ^ { 2 } + 1 } } d x$ can be solved by recognizing it as a standard form related to the inverse hyperbolic sine function, $\sinh^{-1}(x)$. The correct answer involves identifying the coefficient in front of $x$ inside the $\sinh^{-1}$ function and adjusting for the derivative of the inside function, which leads to the answer $\frac { \sinh ^ { - 1 } ( 4 x ) } { 4 } + C$.

The Integral ∫116x2+1dx\int \frac { 1 } { \sqrt { 16 x ^ { 2 } + 1 } } d x∫16x2+1​1​dx

The Integral $\int \frac { 1 } { \sqrt { 16 x ^ { 2 } + 1 } } d x$