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

The integral of (3x^4-4x^3-x-4) is (3x^5)/5-x^4-(x^2)/2-4x+C

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

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

\frac{3 x ^{5}}{5} - x^{4} - \frac{x ^{2}}{2} - 4 x + C

Solution steps

\int 3 x^{4} - 4 x^{3} - x - 4 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 3 x^{4} d x - \int 4 x^{3} d x - \int x d x - \int 4 d x

\int 3 x^{4} d x = \frac{3 x ^{5}}{5}

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

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

\int 4 d x = 4 x

= \frac{3 x ^{5}}{5} - x^{4} - \frac{x ^{2}}{2} - 4 x

Add a constant to the solution

= \frac{3 x ^{5}}{5} - x^{4} - \frac{x ^{2}}{2} - 4 x + C

\int (5 x + 4) d x

\int x^{2} (\frac{x ^{3}}{18} - 1)^{5} d x

\int - 3 z e^{- 3 yz} d y

\int \frac{5}{2 x ^{\frac{9}{4}}} d x

\int 1 + cos (θ) d θ

What is the integral of (3x^4-4x^3-x-4) ?
The integral of (3x^4-4x^3-x-4) is (3x^5)/5-x^4-(x^2)/2-4x+C