Find the most general antiderivative.
-$$\int \left( \frac { 6 } { \sqrt { 1 - x ^ { 2 } } } - \frac { 7 } { x } \right) d x$$
A) $6 \sin ^ { - 1 } x - \ln | x | + C$
B) $6 \sin ^ { - 1 } x + 7 \ln | x | + C$
C) $6 \sin ^ { - 1 } x - 7 \ln | x | + C$
D) $\frac { \sin ^ { - 1 } x } { 6 } - \frac { \ln | x | } { 7 } + C$

Question

Quizplus · Accepted Answer

Use substitution $x=\sin u$ for first part and integration by parts for second part. Thus we get,
\begin{align*}
\int \left( \frac { 6 } { \sqrt { 1 - x ^ { 2 } } } - \frac { 7 } { x } ight) dx &= 6\int \frac{1}{\sqrt{1-\sin^2u}}\cos udu - 7\int\frac{1}{x}dx\
&= 6\int \frac{1}{\cos^2u}d(\sin u)-7\ln |x|\
&= 6\left(	an^{-1}(\sin u)ight)-7\ln |x|+C\
&= 6\left(\sin^{-1}(x)ight)-7\ln |x|+C
\end{align*}
where $C$ is the constant of integration. Thus the correct choice is $\boxed{	extbf{(C) }6 \sin ^ { - 1 } x - 7 \ln | x | + C}$.

Find the Most General Antiderivative ∫(61−x2−7x)dx\int \left( \frac { 6 } { \sqrt { 1 - x ^ { 2 } } } - \frac { 7 } { x } \right) d x∫(1−x2​6​−x7​)dx

Find the Most General Antiderivative $\int \left( \frac { 6 } { \sqrt { 1 - x ^ { 2 } } } - \frac { 7 } { x } \right) d x$