What is the integral of (10)/((x+1)(x^2+9)) ?

The integral of (10)/((x+1)(x^2+9)) is ln|x+1|-1/2 ln|x^2+9|+1/3 arctan(x/3)+C

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integral of (10)/((x+1)(x^2+9))

\int \frac{10}{( x + 1 ) ( x ^{2} + 9 )} d x

ln ∣ x + 1 ∣ - \frac{1}{2} ln x^{2} + 9 + \frac{1}{3} arctan (\frac{x}{3}) + C

Solution steps

\int \frac{10}{( x + 1 ) ( x ^{2} + 9 )} d x

Take the constant out:

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

= 10 \cdot \int \frac{1}{( x + 1 ) ( x ^{2} + 9 )} d x

Take the partial fraction of

\frac{1}{( x + 1 ) ( x ^{2} + 9 )} : \frac{1}{10 ( x + 1 )} + \frac{- x + 1}{10 ( x ^{2} + 9 )}

= 10 \cdot \int \frac{1}{10 ( x + 1 )} + \frac{- x + 1}{10 ( x ^{2} + 9 )} 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

= 10 (\int \frac{1}{10 ( x + 1 )} d x + \int \frac{- x + 1}{10 ( x ^{2} + 9 )} d x)

\int \frac{1}{10 ( x + 1 )} d x = \frac{1}{10} ln ∣ x + 1 ∣

\int \frac{- x + 1}{10 ( x ^{2} + 9 )} d x = \frac{1}{10} (- \frac{1}{2} ln x^{2} + 9 + \frac{1}{3} arctan (\frac{x}{3}))

= 10 (\frac{1}{10} ln ∣ x + 1 ∣ + \frac{1}{10} (- \frac{1}{2} ln x^{2} + 9 + \frac{1}{3} arctan (\frac{x}{3})))

Simplify

10 (\frac{1}{10} ln ∣ x + 1 ∣ + \frac{1}{10} (- \frac{1}{2} ln x^{2} + 9 + \frac{1}{3} arctan (\frac{x}{3}))) : ln ∣ x + 1 ∣ - \frac{1}{2} ln x^{2} + 9 + \frac{1}{3} arctan (\frac{x}{3})

= ln ∣ x + 1 ∣ - \frac{1}{2} ln x^{2} + 9 + \frac{1}{3} arctan (\frac{x}{3})

Add a constant to the solution

= ln ∣ x + 1 ∣ - \frac{1}{2} ln x^{2} + 9 + \frac{1}{3} arctan (\frac{x}{3}) + C

(e^{- x} - e^{2}) (e^{x} + e^{- 5})

(2, 0), (32, - 5)

\int 27 sec^{4} (x) d x

\int \frac{cot ^{5} ( x )}{sin ( x )} d x

x \to - 3 lim (x)

What is the integral of (10)/((x+1)(x^2+9)) ?
The integral of (10)/((x+1)(x^2+9)) is ln|x+1|-1/2 ln|x^2+9|+1/3 arctan(x/3)+C