Find the integral.
-$\int \frac{x d x}{\left(7 x^{2}+3\right)^{5}}$
A) $-\frac{7}{3}\left(7 x^{2}+3\right)^{-4}+C$
B) $-\frac{1}{14}\left(7 x^{2}+3\right)^{-6}+C$
C) $-\frac{7}{3}\left(7 x^{2}+3\right)^{-6}+C$
D) $-\frac{1}{56}\left(7 x^{2}+3\right)^{-4}+C$

Question

Quizplus · Accepted Answer

Using substitution $ u = 7x^2 + 3 $, $ du = 14x \, dx $, the integral becomes $\int \frac{1}{14} u^{-5} \, du$, which evaluates to $-\frac{1}{56} u^{-4} + C$. Substituting back gives $-\frac{1}{56} (7x^2 + 3)^{-4} + C$.

Find the Integral ∫xdx(7x2+3)5\int \frac{x d x}{\left(7 x^{2}+3\right)^{5}}∫(7x2+3)5xdx​ A) −73(7x2+3)−4+C-\frac{7}{3}\left(7 x^{2}+3\right)^{-4}+C−37​(7x2+3)−4+C B) −114(7x2+3)−6+C-\frac{1}{14}\left(7 x^{2}+3\right)^{-6}+C−141​(7x2+3)−6+C

Find the Integral $\int \frac{x d x}{\left(7 x^{2}+3\right)^{5}}$ A) $-\frac{7}{3}\left(7 x^{2}+3\right)^{-4}+C$ B) $-\frac{1}{14}\left(7 x^{2}+3\right)^{-6}+C$