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

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

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

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

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

Solution steps

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

Expand

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

\int \frac{3 x ^{2}}{( 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)))

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

Simplify

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

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

Add a constant to the solution

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

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