What is the integral of (3u^{-2}-4u^2+1) ?

The integral of (3u^{-2}-4u^2+1) is -3/u-(4u^3)/3+u+C

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integral of (3u^{-2}-4u^2+1)

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

- \frac{3}{u} - \frac{4 u ^{3}}{3} + u + C

Solution steps

\int 3 u^{- 2} - 4 u^{2} + 1 d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 3 u^{- 2} d u - \int 4 u^{2} d u + \int 1 d u

\int 3 u^{- 2} d u = - \frac{3}{u}

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

\int 1 d u = u

= - \frac{3}{u} - \frac{4 u ^{3}}{3} + u

Add a constant to the solution

= - \frac{3}{u} - \frac{4 u ^{3}}{3} + u + C

\int (3 x - 2)^{20} d x

f (x) = - 6 x, at x = 4

\int 3 sin^{4} (x) d x

x^{4} - 3 x^{2} + 2

\frac{d}{d x} (cos (x^{2} + 2 x))

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