What is the integral of (2x^2+8+6/(x^2+1)) ?

The integral of (2x^2+8+6/(x^2+1)) is (2x^3)/3+8x+6arctan(x)+C

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integral of (2x^2+8+6/(x^2+1))

\int (2 x^{2} + 8 + \frac{6}{x ^{2} + 1}) d x

\frac{2 x ^{3}}{3} + 8 x + 6 arctan (x) + C

Solution steps

\int 2 x^{2} + 8 + \frac{6}{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 2 x^{2} d x + \int 8 d x + \int \frac{6}{x ^{2} + 1} d x

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

\int 8 d x = 8 x

\int \frac{6}{x ^{2} + 1} d x = 6 arctan (x)

= \frac{2 x ^{3}}{3} + 8 x + 6 arctan (x)

Add a constant to the solution

= \frac{2 x ^{3}}{3} + 8 x + 6 arctan (x) + C

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What is the integral of (2x^2+8+6/(x^2+1)) ?
The integral of (2x^2+8+6/(x^2+1)) is (2x^3)/3+8x+6arctan(x)+C