What is the integral of 1/((4x^2-1)(x+7)) ?

The integral of 1/((4x^2-1)(x+7)) is 1/195 ln|2x+14|-1/26 ln|2x+1|+1/30 ln|2x-1|+C

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integral of 1/((4x^2-1)(x+7))

\int \frac{1}{( 4 x ^{2} - 1 ) ( x + 7 )} d x

\frac{1}{195} ln ∣ 2 x + 14 ∣ - \frac{1}{26} ln ∣ 2 x + 1 ∣ + \frac{1}{30} ln ∣ 2 x - 1 ∣ + C

Solution steps

\int \frac{1}{( 4 x ^{2} - 1 ) ( x + 7 )} d x

Apply Integral Substitution

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

Take the partial fraction of

\frac{1}{( u + 14 ) ( u ^{2} - 1 )} : \frac{1}{195 ( u + 14 )} - \frac{1}{26 ( u + 1 )} + \frac{1}{30 ( u - 1 )}

= \int \frac{1}{195 ( u + 14 )} - \frac{1}{26 ( u + 1 )} + \frac{1}{30 ( u - 1 )} d u

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{1}{195 ( u + 14 )} d u - \int \frac{1}{26 ( u + 1 )} d u + \int \frac{1}{30 ( u - 1 )} d u

\int \frac{1}{195 ( u + 14 )} d u = \frac{1}{195} ln ∣ u + 14 ∣

\int \frac{1}{26 ( u + 1 )} d u = \frac{1}{26} ln ∣ u + 1 ∣

\int \frac{1}{30 ( u - 1 )} d u = \frac{1}{30} ln ∣ u - 1 ∣

= \frac{1}{195} ln ∣ u + 14 ∣ - \frac{1}{26} ln ∣ u + 1 ∣ + \frac{1}{30} ln ∣ u - 1 ∣

Substitute back

u = 2 x

= \frac{1}{195} ln ∣ 2 x + 14 ∣ - \frac{1}{26} ln ∣ 2 x + 1 ∣ + \frac{1}{30} ln ∣ 2 x - 1 ∣

Add a constant to the solution

= \frac{1}{195} ln ∣ 2 x + 14 ∣ - \frac{1}{26} ln ∣ 2 x + 1 ∣ + \frac{1}{30} ln ∣ 2 x - 1 ∣ + C

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What is the integral of 1/((4x^2-1)(x+7)) ?
The integral of 1/((4x^2-1)(x+7)) is 1/195 ln|2x+14|-1/26 ln|2x+1|+1/30 ln|2x-1|+C