What is the integral of (2x-3)^{-1}(x+1)^{-1} ?

The integral of (2x-3)^{-1}(x+1)^{-1} is 1/5 ln|2x-3|-1/5 ln|x+1|+C

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integral of (2x-3)^{-1}(x+1)^{-1}

\int (2 x - 3)^{- 1} (x + 1)^{- 1} d x

\frac{1}{5} ln ∣ 2 x - 3 ∣ - \frac{1}{5} ln ∣ x + 1 ∣ + C

Solution steps

\int (2 x - 3)^{- 1} (x + 1)^{- 1} d x

Simplify

(2 x - 3)^{- 1} (x + 1)^{- 1} : \frac{1}{( 2 x - 3 ) ( x + 1 )}

= \int \frac{1}{( 2 x - 3 ) ( x + 1 )} d x

Take the partial fraction of

\frac{1}{( 2 x - 3 ) ( x + 1 )} : \frac{2}{5 ( 2 x - 3 )} - \frac{1}{5 ( x + 1 )}

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

\int \frac{2}{5 ( 2 x - 3 )} d x = \frac{1}{5} ln ∣ 2 x - 3 ∣

\int \frac{1}{5 ( x + 1 )} d x = \frac{1}{5} ln ∣ x + 1 ∣

= \frac{1}{5} ln ∣ 2 x - 3 ∣ - \frac{1}{5} ln ∣ x + 1 ∣

Add a constant to the solution

= \frac{1}{5} ln ∣ 2 x - 3 ∣ - \frac{1}{5} ln ∣ x + 1 ∣ + C

3 x^{3} + 9 x y^{2} = 5 x^{3}

\frac{d}{d x} (e^{\frac{( x ^{2} )}{2}})

\int (5 x - 2)^{3} d x

y^{^{''}} - 4 y^{^{'}} + 4 y = t^{3} y^{0}

\frac{d ^{2}}{d x ^{2}} (e^{3 x} cos (2 x))

What is the integral of (2x-3)^{-1}(x+1)^{-1} ?
The integral of (2x-3)^{-1}(x+1)^{-1} is 1/5 ln|2x-3|-1/5 ln|x+1|+C