What is the integral of ((26))/((x-1)(x^2+25)) ?

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

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integral of ((26))/((x-1)(x^2+25))

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

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

Solution steps

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

Take the constant out:

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

= 26 \cdot \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 )}

= 26 \cdot \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

= 26 (\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}))

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

Simplify

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

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

Add a constant to the solution

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

\int x^{- 2} e^{\frac{1}{x}} d x

3 x - 8

(5 y - 2 x) y^{^{'}} - 2 y = 0

f (x) = x x^{2} + 60, at x = 2

\frac{\partial}{\partial x} ((3 - x) (3 - y) (x + y - 3))

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