What is the integral of 1/((x+7)(x^2+16)) ?

The integral of 1/((x+7)(x^2+16)) is 1/65 ln|x+7|+1/260 (-2ln|(x^2)/(16)+1|+7arctan(x/4))+C

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integral of 1/((x+7)(x^2+16))

\int \frac{1}{( x + 7 ) ( x ^{2} + 16 )} d x

\frac{1}{65} ln ∣ x + 7 ∣ + \frac{1}{260} (- 2 ln \frac{x ^{2}}{16} + 1 + 7 arctan (\frac{x}{4})) + C

Solution steps

\int \frac{1}{( x + 7 ) ( x ^{2} + 16 )} d x

Take the partial fraction of

\frac{1}{( x + 7 ) ( x ^{2} + 16 )} : \frac{1}{65 ( x + 7 )} + \frac{- x + 7}{65 ( x ^{2} + 16 )}

= \int \frac{1}{65 ( x + 7 )} + \frac{- x + 7}{65 ( x ^{2} + 16 )} 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{1}{65 ( x + 7 )} d x + \int \frac{- x + 7}{65 ( x ^{2} + 16 )} d x

\int \frac{1}{65 ( x + 7 )} d x = \frac{1}{65} ln ∣ x + 7 ∣

\int \frac{- x + 7}{65 ( x ^{2} + 16 )} d x = \frac{1}{260} (- 2 ln \frac{x ^{2}}{16} + 1 + 7 arctan (\frac{x}{4}))

= \frac{1}{65} ln ∣ x + 7 ∣ + \frac{1}{260} (- 2 ln \frac{x ^{2}}{16} + 1 + 7 arctan (\frac{x}{4}))

Add a constant to the solution

= \frac{1}{65} ln ∣ x + 7 ∣ + \frac{1}{260} (- 2 ln \frac{x ^{2}}{16} + 1 + 7 arctan (\frac{x}{4})) + C

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1 - 9 x

\int_{0}^{4} 0.125 x^{3} d x

e^{- x} arctan (2 e^{x})

\int (3 x^{2} + 4 x + 1) d x

What is the integral of 1/((x+7)(x^2+16)) ?
The integral of 1/((x+7)(x^2+16)) is 1/65 ln|x+7|+1/260 (-2ln|(x^2)/(16)+1|+7arctan(x/4))+C