What is the integral of (x^7-3x^4+x) ?

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

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integral of (x^7-3x^4+x)

\int (x^{7} - 3 x^{4} + x) d x

\frac{x ^{8}}{8} - \frac{3 x ^{5}}{5} + \frac{x ^{2}}{2} + C

Solution steps

\int x^{7} - 3 x^{4} + x 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^{7} d x - \int 3 x^{4} d x + \int x d x

\int x^{7} d x = \frac{x ^{8}}{8}

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

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

= \frac{x ^{8}}{8} - \frac{3 x ^{5}}{5} + \frac{x ^{2}}{2}

Add a constant to the solution

= \frac{x ^{8}}{8} - \frac{3 x ^{5}}{5} + \frac{x ^{2}}{2} + C

\int \frac{1}{( y ^{2} + x ^{2} ) ^{\frac{3}{2}}} d x

\int 7 cos^{5} (x) sin^{4} (x) d x

x \to a lim (3)

d er i v a t i v e y = ln (ln (5 x))

\frac{d}{d x} (\frac{15}{x})

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