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

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

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

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

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

Solution steps

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

\int 2 x^{5} d x = \frac{x ^{6}}{3}

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

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

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

Add a constant to the solution

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

\int \frac{sec ( x )}{cos ( x )} d x

\frac{\partial}{\partial x} (49 - 7 x^{2} - y^{2})

\int \frac{x ^{3}}{( 1 + x ^{4} ) ^{\frac{1}{3}}} d x

\int 12 x + 4 y d y

x \to 2 lim (a x + 7 + b)

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