What is the integral of 4x^2+3+8/(x^2+1) ?

The integral of 4x^2+3+8/(x^2+1) is (4x^3)/3+3x+8arctan(x)+C

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integral of 4x^2+3+8/(x^2+1)

\int 4 x^{2} + 3 + \frac{8}{x ^{2} + 1} d x

\frac{4 x ^{3}}{3} + 3 x + 8 arctan (x) + C

Solution steps

\int 4 x^{2} + 3 + \frac{8}{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 4 x^{2} d x + \int 3 d x + \int \frac{8}{x ^{2} + 1} d x

\int 4 x^{2} d x = \frac{4 x ^{3}}{3}

\int 3 d x = 3 x

\int \frac{8}{x ^{2} + 1} d x = 8 arctan (x)

= \frac{4 x ^{3}}{3} + 3 x + 8 arctan (x)

Add a constant to the solution

= \frac{4 x ^{3}}{3} + 3 x + 8 arctan (x) + C

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What is the integral of 4x^2+3+8/(x^2+1) ?
The integral of 4x^2+3+8/(x^2+1) is (4x^3)/3+3x+8arctan(x)+C