What is the integral of (1+6x)/((4x-3)(2x+5)) ?

The integral of (1+6x)/((4x-3)(2x+5)) is 11/52 ln|4x-3|+7/13 ln|2x+5|+C

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integral of (1+6x)/((4x-3)(2x+5))

\int \frac{1 + 6 x}{( 4 x - 3 ) ( 2 x + 5 )} d x

\frac{11}{52} ln ∣ 4 x - 3 ∣ + \frac{7}{13} ln ∣ 2 x + 5 ∣ + C

Solution steps

\int \frac{1 + 6 x}{( 4 x - 3 ) ( 2 x + 5 )} d x

Take the partial fraction of

\frac{1 + 6 x}{( 4 x - 3 ) ( 2 x + 5 )} : \frac{11}{13 ( 4 x - 3 )} + \frac{14}{13 ( 2 x + 5 )}

= \int \frac{11}{13 ( 4 x - 3 )} + \frac{14}{13 ( 2 x + 5 )} d x

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{11}{13 ( 4 x - 3 )} d x + \int \frac{14}{13 ( 2 x + 5 )} d x

\int \frac{11}{13 ( 4 x - 3 )} d x = \frac{11}{52} ln ∣ 4 x - 3 ∣

\int \frac{14}{13 ( 2 x + 5 )} d x = \frac{7}{13} ln ∣ 2 x + 5 ∣

= \frac{11}{52} ln ∣ 4 x - 3 ∣ + \frac{7}{13} ln ∣ 2 x + 5 ∣

Add a constant to the solution

= \frac{11}{52} ln ∣ 4 x - 3 ∣ + \frac{7}{13} ln ∣ 2 x + 5 ∣ + C

5 x y^{^{'}} = 2 x e^{x} - 5 y + 6 x^{2}

x \to 0 + lim (\frac{7 x}{x})

6 y^{^{''}} + 8 y^{^{'}} + 9 y = 0

y^{^{''}} + 6 y^{^{'}} + 8 y = 3 e^{- 2 x} + 2 x

\frac{\partial}{\partial x} (2 x)

What is the integral of (1+6x)/((4x-3)(2x+5)) ?
The integral of (1+6x)/((4x-3)(2x+5)) is 11/52 ln|4x-3|+7/13 ln|2x+5|+C