What is the integral of x^3-4x^2+3x ?

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

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integral of x^3-4x^2+3x

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

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

Solution steps

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

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

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

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

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

Add a constant to the solution

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

\int \frac{( 2 x ^{2} + x )}{( 4 x ^{3} + 3 x ^{2} ) ^{2}} d x

\int cos (d) t

\int tan (x) d (sec (x)) d x

\int \frac{1}{( arctan ( \frac{y}{9} ) ) ( 81 + y ^{2} )} d y

\int (x^{2} - 4)^{3} d x

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