What is the integral of (3-10x)/(4x^2+9) ?

The integral of (3-10x)/(4x^2+9) is 1/2 arctan(2/3 x)-5/4 ln|4x^2+9|+C

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integral of (3-10x)/(4x^2+9)

\int \frac{3 - 10 x}{4 x ^{2} + 9} d x

\frac{1}{2} arctan (\frac{2}{3} x) - \frac{5}{4} ln 4 x^{2} + 9 + C

Solution steps

\int \frac{3 - 10 x}{4 x ^{2} + 9} d x

Expand

\frac{3 - 10 x}{4 x ^{2} + 9} : \frac{3}{4 x ^{2} + 9} - \frac{10 x}{4 x ^{2} + 9}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{3}{4 x ^{2} + 9} d x - \int \frac{10 x}{4 x ^{2} + 9} d x

\int \frac{3}{4 x ^{2} + 9} d x = \frac{1}{2} arctan (\frac{2}{3} x)

\int \frac{10 x}{4 x ^{2} + 9} d x = \frac{5}{4} ln 4 x^{2} + 9

= \frac{1}{2} arctan (\frac{2}{3} x) - \frac{5}{4} ln 4 x^{2} + 9

Add a constant to the solution

= \frac{1}{2} arctan (\frac{2}{3} x) - \frac{5}{4} ln 4 x^{2} + 9 + C

What is the integral of (3-10x)/(4x^2+9) ?
The integral of (3-10x)/(4x^2+9) is 1/2 arctan(2/3 x)-5/4 ln|4x^2+9|+C