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

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

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

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

2 x^{2} - 3 x + 3 arctan (x) + 2 + C

Solution steps

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

Expand

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

\int \frac{4 x ^{3}}{x ^{2} + 1} d x = 2 (x^{2} + 1 - ln x^{2} + 1)

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

\int \frac{4 x}{x ^{2} + 1} d x = 2 ln x^{2} + 1

= 2 (x^{2} + 1 - ln x^{2} + 1) - 3 (- arctan (x) + x) + 2 ln x^{2} + 1

Simplify

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

= 2 x^{2} - 3 x + 3 arctan (x) + 2

Add a constant to the solution

= 2 x^{2} - 3 x + 3 arctan (x) + 2 + C

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