# Question #4d270

$\int \frac{3 x}{2 {x}^{2} + 9} \mathrm{dx} = \frac{3}{4} \cdot \ln \left(2 {x}^{2} + 9\right) + C$

#### Explanation:

There is no need to reduction to partial fraction. The numerator has the differential of the denominator.

$\int \frac{3 x}{2 {x}^{2} + 9} \mathrm{dx} = \frac{3}{4} \int \frac{4 x}{2 {x}^{2} + 9} \mathrm{dx}$

At this point, we can use the formula $\int \frac{\mathrm{du}}{u} = \ln u + C$

$\int \frac{3 x}{2 {x}^{2} + 9} \mathrm{dx} = \frac{3}{4} \cdot \ln \left(2 {x}^{2} + 9\right) + C$

God bless....I hope the explanation is useful.

May 6, 2016

$\frac{3}{4} \log \left(2 {x}^{2} + 9\right)$

#### Explanation:

$\setminus \int 3 \frac{x}{2 {x}^{2} + 9} \mathrm{dx} = \frac{3}{4} \setminus \int 4 x \frac{\mathrm{dx}}{2 {x}^{2} + 9} = \frac{3}{4} \setminus \int \frac{d \left(2 {x}^{2} + 9\right)}{2 {x}^{2} + 9} = \frac{3}{4} \log \left(2 {x}^{2} + 9\right)$