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

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

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integral of (x^2-1)(4+3x)

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

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

Solution steps

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

Expand

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

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

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

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

\int 4 d x = 4 x

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

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

Add a constant to the solution

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

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