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

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

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

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

\frac{3 x ^{2}}{2} - 4 x + 4 arctan (x) + C

Solution steps

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

Long division

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

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

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

\int 4 d x = 4 x

\int \frac{4}{x ^{2} + 1} d x = 4 arctan (x)

= \frac{3 x ^{2}}{2} - 4 x + 4 arctan (x)

Add a constant to the solution

= \frac{3 x ^{2}}{2} - 4 x + 4 arctan (x) + C

\int 6 x^{5} - 7 x^{4} - 9 x^{2} d x

\int x + c d x

\int_{0}^{1} \int_{0}^{x} (\frac{6}{7} (x^{2} + x \cdot \frac{y}{2})) d y d x

y = \frac{t}{( t - 7 ) ^{2}}

x \to \infty lim (ln (9 x) - ln (x + 7))

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