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

The integral of (x+1)(5x^3-2) is x^5-x^2+(5x^4)/4-2x+C

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integral of (x+1)(5x^3-2)

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

x^{5} - x^{2} + \frac{5 x ^{4}}{4} - 2 x + C

Solution steps

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

Expand

(x + 1) (5 x^{3} - 2) : 5 x^{4} - 2 x + 5 x^{3} - 2

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

\int 5 x^{4} d x = x^{5}

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

\int 5 x^{3} d x = \frac{5 x ^{4}}{4}

\int 2 d x = 2 x

= x^{5} - x^{2} + \frac{5 x ^{4}}{4} - 2 x

Add a constant to the solution

= x^{5} - x^{2} + \frac{5 x ^{4}}{4} - 2 x + C

7 - x

sin^{5} (x)

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

(2 x + 3) y^{^{'}} = y + 2 x + 3

\frac{1 d y}{x ^{2} d x} = y^{2} + 9

What is the integral of (x+1)(5x^3-2) ?
The integral of (x+1)(5x^3-2) is x^5-x^2+(5x^4)/4-2x+C