What is the integral of (3x-1)(x+5) ?

The integral of (3x-1)(x+5) is x^3+7x^2-5x+C

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integral of (3x-1)(x+5)

\int (3 x - 1) (x + 5) d x

x^{3} + 7 x^{2} - 5 x + C

Solution steps

\int (3 x - 1) (x + 5) d x

Expand

(3 x - 1) (x + 5) : 3 x^{2} + 14 x - 5

= \int 3 x^{2} + 14 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 14 x d x - \int 5 d x

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

\int 14 x d x = 7 x^{2}

\int 5 d x = 5 x

= x^{3} + 7 x^{2} - 5 x

Add a constant to the solution

= x^{3} + 7 x^{2} - 5 x + C

f (x) = (3 + x^{5})^{3} - (2 + x^{3})^{4}

\frac{d}{d x} (arccos (\frac{4}{x}))

\int 7 x d x

y^{^{''}} - 2 y^{^{'}} + 1 = 0

\int 0. 4^{\frac{x}{2}} d x

What is the integral of (3x-1)(x+5) ?
The integral of (3x-1)(x+5) is x^3+7x^2-5x+C