Evaluate the definite integral.
-$\int_{0}^{1} \frac{x}{\left(3 x^{2}+2\right)^{3}} d x$
A) 0.0175
B) -0.0208
C) 0.0032
D) 0.1050

Question

Quizplus · Accepted Answer

Using the substitution $u = 3x^2 + 2$, $du = 6x dx$, we get$\int_{0}^{1} \frac{x}{\left(3 x^{2}+2\right)^{3}} d x = \int_{2}^{5} \frac{1}{6u^{3}} du = \left[-\frac{1}{12u^{2}}\right]_{2}^{5} = -\frac{1}{12(5)^{2}} + \frac{1}{12(2)^{2}} = 0.0175$

Evaluate the Definite Integral ∫01x(3x2+2)3dx\int_{0}^{1} \frac{x}{\left(3 x^{2}+2\right)^{3}} d x∫01​(3x2+2)3x​dx A) 00175 B) -0

Evaluate the Definite Integral $\int_{0}^{1} \frac{x}{\left(3 x^{2}+2\right)^{3}} d x$ A) 00175
B) -0