What is the integral of x^5+8x^3-9x^2+7x-5 ?

The integral of x^5+8x^3-9x^2+7x-5 is (x^6)/6+2x^4-3x^3+(7x^2)/2-5x+C

Popular >

integral of x^5+8x^3-9x^2+7x-5

\int x^{5} + 8 x^{3} - 9 x^{2} + 7 x - 5 d x

\frac{x ^{6}}{6} + 2 x^{4} - 3 x^{3} + \frac{7 x ^{2}}{2} - 5 x + C

Solution steps

\int x^{5} + 8 x^{3} - 9 x^{2} + 7 x - 5 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 x^{5} d x + \int 8 x^{3} d x - \int 9 x^{2} d x + \int 7 x d x - \int 5 d x

\int x^{5} d x = \frac{x ^{6}}{6}

\int 8 x^{3} d x = 2 x^{4}

\int 9 x^{2} d x = 3 x^{3}

\int 7 x d x = \frac{7 x ^{2}}{2}

\int 5 d x = 5 x

= \frac{x ^{6}}{6} + 2 x^{4} - 3 x^{3} + \frac{7 x ^{2}}{2} - 5 x

Add a constant to the solution

= \frac{x ^{6}}{6} + 2 x^{4} - 3 x^{3} + \frac{7 x ^{2}}{2} - 5 x + C

\int - x^{3} + \frac{1}{x ^{2}} d x

\int \frac{4 e ^{- \frac{3}{x ^{5}}}}{x ^{6}} d x

\frac{\partial}{\partial y} ((x + y) e^{y^{2}})

3 t + 4

f (x) = x^{3} + x^{2} + x + 1

What is the integral of x^5+8x^3-9x^2+7x-5 ?
The integral of x^5+8x^3-9x^2+7x-5 is (x^6)/6+2x^4-3x^3+(7x^2)/2-5x+C