Question #4d270

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Question #4d270

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Answer 1 · 2016-05-06T05:44:32

$\int \frac{3 x}{2 x^{2} + 9} d x = \frac{3}{4} \cdot ln (2 x^{2} + 9) + C$ $int (3x)/(2x^2+9) dx=3/4*ln(2x^2+9)+C$

Explanation:

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

$\int \frac{3 x}{2 x^{2} + 9} d x = \frac{3}{4} \int \frac{4 x}{2 x^{2} + 9} d x$ $int (3x)/(2x^2+9) dx=3/4int (4x)/(2x^2+9) dx$

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

$\int \frac{3 x}{2 x^{2} + 9} d x = \frac{3}{4} \cdot ln (2 x^{2} + 9) + C$ $int (3x)/(2x^2+9) dx=3/4*ln(2x^2+9)+C$

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

Answer 2 · 2016-05-06T05:47:02

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

Explanation:

$\int 3 \frac{x}{2 x^{2} + 9} d x = \frac{3}{4} \int 4 x \frac{d x}{2 x^{2} + 9} = \frac{3}{4} \int \frac{d (2 x^{2} + 9)}{2 x^{2} + 9} = \frac{3}{4} log (2 x^{2} + 9)$ $\int 3x/(2x^2+9) dx = 3/4 \int 4x dx/(2x^2 + 9) = 3/4 \int (d(2x^2 + 9))/(2x^2 + 9) = 3/4 log(2x^2 + 9)$

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Question #4d270

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