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

The integral of (2x^2+3x-2)(4x+3) is 2x^4+6x^3+1/2 x^2-6x+C

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

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

2 x^{4} + 6 x^{3} + \frac{1}{2} x^{2} - 6 x + C

Solution steps

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

Apply Integration By Parts

: (2 x^{2} + 3 x - 2) (4 x + 3) x - \int (24 x^{2} + 36 x + 1) x d x

= (2 x^{2} + 3 x - 2) (4 x + 3) x - \int (24 x^{2} + 36 x + 1) x d x

\int (24 x^{2} + 36 x + 1) x d x = 6 x^{4} + 12 x^{3} + \frac{1}{2} x^{2}

= (2 x^{2} + 3 x - 2) (4 x + 3) x - (6 x^{4} + 12 x^{3} + \frac{1}{2} x^{2})

Simplify

(2 x^{2} + 3 x - 2) (4 x + 3) x - (6 x^{4} + 12 x^{3} + \frac{1}{2} x^{2}) : 2 x^{4} + 6 x^{3} + \frac{1}{2} x^{2} - 6 x

= 2 x^{4} + 6 x^{3} + \frac{1}{2} x^{2} - 6 x

Add a constant to the solution

= 2 x^{4} + 6 x^{3} + \frac{1}{2} x^{2} - 6 x + C

\int_{1}^{2} 12 x^{2} 3^{x^{3}} d x

\int \frac{( a + b x ^{2} )}{3 a x + b x ^{3}} d x

\int_{- 1}^{0} y + 1 d y

y^{^{'}} = 2 t, y (0) = 0

\int \frac{1}{486} (81 - x^{2}) d x

What is the integral of (2x^2+3x-2)(4x+3) ?
The integral of (2x^2+3x-2)(4x+3) is 2x^4+6x^3+1/2 x^2-6x+C