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

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

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

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

x^{3} - 2 x^{2} - 5 x + C

Solution steps

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

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

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

\int 5 d x = 5 x

= x^{3} - 2 x^{2} - 5 x

Add a constant to the solution

= x^{3} - 2 x^{2} - 5 x + C

\int \frac{2 x}{6 x - x ^{2}} d x

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\int \frac{3 x ^{4}}{1 - 6 x ^{5}} d x

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

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