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

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

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

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

3 x + \frac{3}{2} ln x^{2} + 1 + C

Solution steps

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

Long division

\frac{3 x ^{2} + 3 x + 3}{x ^{2} + 1} : 3 + \frac{3 x}{x ^{2} + 1}

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

\int 3 d x = 3 x

\int \frac{3 x}{x ^{2} + 1} d x = \frac{3}{2} ln x^{2} + 1

= 3 x + \frac{3}{2} ln x^{2} + 1

Add a constant to the solution

= 3 x + \frac{3}{2} ln x^{2} + 1 + C

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