What is the integral of (7x)/((2x-3)(x+2)) ?

The integral of (7x)/((2x-3)(x+2)) is 3/2 ln|2x-3|+2ln|x+2|+C

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integral of (7x)/((2x-3)(x+2))

\int \frac{7 x}{( 2 x - 3 ) ( x + 2 )} d x

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

Solution steps

\int \frac{7 x}{( 2 x - 3 ) ( x + 2 )} d x

Take the constant out:

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

= 7 \cdot \int \frac{x}{( 2 x - 3 ) ( x + 2 )} d x

Take the partial fraction of

\frac{x}{( 2 x - 3 ) ( x + 2 )} : \frac{3}{7 ( 2 x - 3 )} + \frac{2}{7 ( x + 2 )}

= 7 \cdot \int \frac{3}{7 ( 2 x - 3 )} + \frac{2}{7 ( x + 2 )} 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

= 7 (\int \frac{3}{7 ( 2 x - 3 )} d x + \int \frac{2}{7 ( x + 2 )} d x)

\int \frac{3}{7 ( 2 x - 3 )} d x = \frac{3}{14} ln ∣ 2 x - 3 ∣

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

= 7 (\frac{3}{14} ln ∣ 2 x - 3 ∣ + \frac{2}{7} ln ∣ x + 2 ∣)

Simplify

7 (\frac{3}{14} ln ∣ 2 x - 3 ∣ + \frac{2}{7} ln ∣ x + 2 ∣) : \frac{3}{2} ln ∣ 2 x - 3 ∣ + 2 ln ∣ x + 2 ∣

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

Add a constant to the solution

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

\int tan^{9} (t) sec^{4} (t) d t

\frac{4}{( 4 - x ^{6} ) ^{\frac{3}{2}}}

x y^{^{'}} + y = - 2 x y^{2}

cos (e^{- 2 θ^{3}})

\frac{d}{d x} (4 x^{- \frac{3}{8}})

What is the integral of (7x)/((2x-3)(x+2)) ?
The integral of (7x)/((2x-3)(x+2)) is 3/2 ln|2x-3|+2ln|x+2|+C