What is the integral of (5x+3)/((x+1)(x^2+1)) ?

The integral of (5x+3)/((x+1)(x^2+1)) is -ln|x+1|+1/2 ln|x^2+1|+4arctan(x)+C

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integral of (5x+3)/((x+1)(x^2+1))

\int \frac{5 x + 3}{( x + 1 ) ( x ^{2} + 1 )} d x

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

Solution steps

\int \frac{5 x + 3}{( x + 1 ) ( x ^{2} + 1 )} d x

Expand

\frac{5 x + 3}{( x + 1 ) ( x ^{2} + 1 )} : \frac{5 x}{( x + 1 ) ( x ^{2} + 1 )} + \frac{3}{( x + 1 ) ( x ^{2} + 1 )}

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

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

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

= 5 (- \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} (\frac{1}{2} ln x^{2} + 1 + arctan (x))) + 3 (\frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} (- \frac{1}{2} ln x^{2} + 1 + arctan (x)))

Simplify

5 (- \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} (\frac{1}{2} ln x^{2} + 1 + arctan (x))) + 3 (\frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} (- \frac{1}{2} ln x^{2} + 1 + arctan (x))) : - ln ∣ x + 1 ∣ + \frac{1}{2} ln x^{2} + 1 + 4 arctan (x)

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

Add a constant to the solution

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

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What is the integral of (5x+3)/((x+1)(x^2+1)) ?
The integral of (5x+3)/((x+1)(x^2+1)) is -ln|x+1|+1/2 ln|x^2+1|+4arctan(x)+C