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Answer 1 · 2016-05-06T05:44:32

$int (3x)/(2x^2+9) dx=3/4*ln(2x^2+9)+C$

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

$int (3x)/(2x^2+9) dx=3/4int (4x)/(2x^2+9) dx$

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

$int (3x)/(2x^2+9) dx=3/4*ln(2x^2+9)+C$

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

Answer 2 · 2016-05-06T05:47:02

$3/4 log(2x^2 + 9)$

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