What is the integral of (7x-5)/(9x^2+4) ?

The integral of (7x-5)/(9x^2+4) is 1/18 (7ln| 9/4 x^2+1|-15arctan(3/2 x))+C

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integral of (7x-5)/(9x^2+4)

\int \frac{7 x - 5}{9 x ^{2} + 4} d x

\frac{1}{18} (7 ln \frac{9}{4} x^{2} + 1 - 15 arctan (\frac{3}{2} x)) + C

Solution steps

\int \frac{7 x - 5}{9 x ^{2} + 4} d x

Apply Integral Substitution

= \int \frac{14 u - 15}{18 ( u ^{2} + 1 )} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{18} \cdot \int \frac{14 u - 15}{u ^{2} + 1} d u

Expand

\frac{14 u - 15}{u ^{2} + 1} : \frac{14 u}{u ^{2} + 1} - \frac{15}{u ^{2} + 1}

Apply the Sum Rule:

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

= \frac{1}{18} (\int \frac{14 u}{u ^{2} + 1} d u - \int \frac{15}{u ^{2} + 1} d u)

\int \frac{14 u}{u ^{2} + 1} d u = 7 ln u^{2} + 1

\int \frac{15}{u ^{2} + 1} d u = 15 arctan (u)

= \frac{1}{18} (7 ln u^{2} + 1 - 15 arctan (u))

Substitute back

u = \frac{3}{2} x

= \frac{1}{18} (7 ln (\frac{3}{2} x)^{2} + 1 - 15 arctan (\frac{3}{2} x))

(\frac{3}{2} x)^{2} = \frac{9}{4} x^{2}

= \frac{1}{18} (7 ln \frac{9}{4} x^{2} + 1 - 15 arctan (\frac{3}{2} x))

Add a constant to the solution

= \frac{1}{18} (7 ln \frac{9}{4} x^{2} + 1 - 15 arctan (\frac{3}{2} x)) + C

What is the integral of (7x-5)/(9x^2+4) ?
The integral of (7x-5)/(9x^2+4) is 1/18 (7ln| 9/4 x^2+1|-15arctan(3/2 x))+C