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

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

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

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

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

Solution steps

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

Expand

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

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

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 t^{2} d t + \int 3 t^{3} d t - \int 4 d t - \int 3 t d t

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

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

\int 4 d t = 4 t

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

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

Add a constant to the solution

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

f (x) = x^{4}, at x = 1

y = \frac{6}{x}, (25, \frac{6}{5})

\frac{x ^{2}}{7 + x}

x \to \infty lim (\frac{( 2 x ^{2} + 1 )}{3 x - 5})

y^{^{''}} - 2 y^{^{'}} = 2 t

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