What is the integral of 1/((x+1)(x^2+25)) ?

The integral of 1/((x+1)(x^2+25)) is 1/26 ln|x+1|+1/26 (-1/2 ln|x^2+25|+1/5 arctan(x/5))+C

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integral of 1/((x+1)(x^2+25))

\int \frac{1}{( x + 1 ) ( x ^{2} + 25 )} d x

\frac{1}{26} ln ∣ x + 1 ∣ + \frac{1}{26} (- \frac{1}{2} ln x^{2} + 25 + \frac{1}{5} arctan (\frac{x}{5})) + C

Solution steps

\int \frac{1}{( x + 1 ) ( x ^{2} + 25 )} d x

Take the partial fraction of

\frac{1}{( x + 1 ) ( x ^{2} + 25 )} : \frac{1}{26 ( x + 1 )} + \frac{- x + 1}{26 ( x ^{2} + 25 )}

= \int \frac{1}{26 ( x + 1 )} + \frac{- x + 1}{26 ( x ^{2} + 25 )} 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}{26 ( x + 1 )} d x + \int \frac{- x + 1}{26 ( x ^{2} + 25 )} d x

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

\int \frac{- x + 1}{26 ( x ^{2} + 25 )} d x = \frac{1}{26} (- \frac{1}{2} ln x^{2} + 25 + \frac{1}{5} arctan (\frac{x}{5}))

= \frac{1}{26} ln ∣ x + 1 ∣ + \frac{1}{26} (- \frac{1}{2} ln x^{2} + 25 + \frac{1}{5} arctan (\frac{x}{5}))

Add a constant to the solution

= \frac{1}{26} ln ∣ x + 1 ∣ + \frac{1}{26} (- \frac{1}{2} ln x^{2} + 25 + \frac{1}{5} arctan (\frac{x}{5})) + C

\int (x^{\frac{3}{2}} + 8)^{6} x d x

n = 1 \sum \infty (- \frac{10}{11})^{n}

\frac{d}{d x} (\frac{x}{y ^{2}} - \frac{y}{x ^{2}})

f (x) = \frac{1}{1 + x ^{2}}, at x = - 1

\frac{\partial}{\partial θ} (n ln (2 θ))

What is the integral of 1/((x+1)(x^2+25)) ?
The integral of 1/((x+1)(x^2+25)) is 1/26 ln|x+1|+1/26 (-1/2 ln|x^2+25|+1/5 arctan(x/5))+C