What is the integral of (x^5-10x^4+x^3-15x^2) ?

The integral of (x^5-10x^4+x^3-15x^2) is (x^6)/6-2x^5+(x^4)/4-5x^3+C

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integral of (x^5-10x^4+x^3-15x^2)

\int (x^{5} - 10 x^{4} + x^{3} - 15 x^{2}) d x

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

Solution steps

\int x^{5} - 10 x^{4} + x^{3} - 15 x^{2} 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 10 x^{4} d x + \int x^{3} d x - \int 15 x^{2} d x

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

\int 10 x^{4} d x = 2 x^{5}

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

\int 15 x^{2} d x = 5 x^{3}

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

Add a constant to the solution

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

n = 1 \sum \infty \frac{z ^{n}}{n ^{2}}

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

\frac{d y}{d x} = x + 3

x \to 2 lim (4 x)

\int_{1}^{4} (\frac{1}{x})^{2} d x

What is the integral of (x^5-10x^4+x^3-15x^2) ?
The integral of (x^5-10x^4+x^3-15x^2) is (x^6)/6-2x^5+(x^4)/4-5x^3+C