What is the integral of y=[3x+2][2x-5] ?

The integral of y=[3x+2][2x-5] is 2x^3+2x^2-(15x^2)/2-10x+C

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integral of y=[3x+2][2x-5]

integral

y = [3 x + 2] [2 x - 5]

2 x^{3} + 2 x^{2} - \frac{15 x ^{2}}{2} - 10 x + C

Solution steps

\int [3 x + 2] [2 x - 5] d x

Multiply by the conjugate of

3 x + 2 : \frac{9 x ^{2} - 4}{3 x - 2}

= \int [\frac{9 x ^{2} - 4}{3 x - 2}] [2 x - 5] d x

Simplify

[\frac{9 x ^{2} - 4}{3 x - 2}] [2 x - 5] : (3 x + 2) (2 x - 5)

= \int (3 x + 2) (2 x - 5) d x

Apply Integration By Parts

: 3 x^{3} + 4 x^{2} - \frac{15 x ^{2}}{2} - 10 x - \int 3 x^{2} + 4 x d x

= 3 x^{3} + 4 x^{2} - \frac{15 x ^{2}}{2} - 10 x - \int 3 x^{2} + 4 x d x

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

= 3 x^{3} + 4 x^{2} - \frac{15 x ^{2}}{2} - 10 x - (x^{3} + 2 x^{2})

Simplify

3 x^{3} + 4 x^{2} - \frac{15 x ^{2}}{2} - 10 x - (x^{3} + 2 x^{2}) : 2 x^{3} + 2 x^{2} - \frac{15 x ^{2}}{2} - 10 x

= 2 x^{3} + 2 x^{2} - \frac{15 x ^{2}}{2} - 10 x

Add a constant to the solution

= 2 x^{3} + 2 x^{2} - \frac{15 x ^{2}}{2} - 10 x + C

\frac{1}{( k + e ^{- t} )}

e^{\frac{y ^{2}}{2}}

x (x^{2})

\frac{1}{( 1 - 2 x ^{2} )}

y = 4 t + 2

What is the integral of y=[3x+2][2x-5] ?
The integral of y=[3x+2][2x-5] is 2x^3+2x^2-(15x^2)/2-10x+C