What is the integral of x^3-8x^2+16x+1 ?

The integral of x^3-8x^2+16x+1 is (x^4)/4-(8x^3)/3+8x^2+x+C

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integral of x^3-8x^2+16x+1

\int x^{3} - 8 x^{2} + 16 x + 1 d x

\frac{x ^{4}}{4} - \frac{8 x ^{3}}{3} + 8 x^{2} + x + C

Solution steps

\int x^{3} - 8 x^{2} + 16 x + 1 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^{3} d x - \int 8 x^{2} d x + \int 16 x d x + \int 1 d x

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

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

\int 16 x d x = 8 x^{2}

\int 1 d x = x

= \frac{x ^{4}}{4} - \frac{8 x ^{3}}{3} + 8 x^{2} + x

Add a constant to the solution

= \frac{x ^{4}}{4} - \frac{8 x ^{3}}{3} + 8 x^{2} + x + C

3 x^{\frac{1}{3}} - 5 x^{\frac{4}{3}}

\int e^{7 x + 1} d x

\frac{\partial}{\partial x} (\frac{x + 3}{y + 3})

\frac{s}{s + a}

\int \frac{1}{x ^{2} - 2 x + 122} d x

What is the integral of x^3-8x^2+16x+1 ?
The integral of x^3-8x^2+16x+1 is (x^4)/4-(8x^3)/3+8x^2+x+C