What is the integral of (x^3+4x^2-5) ?

The integral of (x^3+4x^2-5) is (x^4)/4+(4x^3)/3-5x+C

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integral of (x^3+4x^2-5)

\int (x^{3} + 4 x^{2} - 5) d x

\frac{x ^{4}}{4} + \frac{4 x ^{3}}{3} - 5 x + C

Solution steps

\int x^{3} + 4 x^{2} - 5 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 5 d x

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

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

\int 5 d x = 5 x

= \frac{x ^{4}}{4} + \frac{4 x ^{3}}{3} - 5 x

Add a constant to the solution

= \frac{x ^{4}}{4} + \frac{4 x ^{3}}{3} - 5 x + C

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

\int 4 + 4 t^{2} + t^{4} d t

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

\int \frac{x + 10}{2 x ^{2} + 5 x - 3} d x

\frac{1}{1 - x}

What is the integral of (x^3+4x^2-5) ?
The integral of (x^3+4x^2-5) is (x^4)/4+(4x^3)/3-5x+C