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

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

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

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

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

Solution steps

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

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

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

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

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

Add a constant to the solution

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

\int \frac{9 x}{1 + x ^{2}} d x

\int \frac{2 x ^{3} + x - 1}{x ^{2} - 5 x} d x

e^{5 x}

\int \frac{e ^{x}}{3 + 2 e ^{x}} d x

\int \frac{( x - 9 ) ^{6}}{2 x} d x

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