What is the integral of x^4-8x^3+26x^2-40x+25 ?

The integral of x^4-8x^3+26x^2-40x+25 is (x^5)/5-2x^4+(26x^3)/3-20x^2+25x+C

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integral of x^4-8x^3+26x^2-40x+25

\int x^{4} - 8 x^{3} + 26 x^{2} - 40 x + 25 d x

\frac{x ^{5}}{5} - 2 x^{4} + \frac{26 x ^{3}}{3} - 20 x^{2} + 25 x + C

Solution steps

\int x^{4} - 8 x^{3} + 26 x^{2} - 40 x + 25 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^{4} d x - \int 8 x^{3} d x + \int 26 x^{2} d x - \int 40 x d x + \int 25 d x

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

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

\int 26 x^{2} d x = \frac{26 x ^{3}}{3}

\int 40 x d x = 20 x^{2}

\int 25 d x = 25 x

= \frac{x ^{5}}{5} - 2 x^{4} + \frac{26 x ^{3}}{3} - 20 x^{2} + 25 x

Add a constant to the solution

= \frac{x ^{5}}{5} - 2 x^{4} + \frac{26 x ^{3}}{3} - 20 x^{2} + 25 x + C

x \to 6 lim (2 x - 3)

x \to 7 lim (\frac{x - 7}{x - 7})

\frac{d}{d x} (x^{3} - 5 e^{2 x})

\frac{d}{d x} (e^{- e^{- x}})

x \to 0.5 + lim (3 x)

What is the integral of x^4-8x^3+26x^2-40x+25 ?
The integral of x^4-8x^3+26x^2-40x+25 is (x^5)/5-2x^4+(26x^3)/3-20x^2+25x+C