What is the integral of (8x^2+3x^4+5/(2x^8)+e^x) ?

The integral of (8x^2+3x^4+5/(2x^8)+e^x) is (8x^3)/3+(3x^5)/5-5/(14x^7)+e^x+C

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integral of (8x^2+3x^4+5/(2x^8)+e^x)

\int (8 x^{2} + 3 x^{4} + \frac{5}{2 x ^{8}} + e^{x}) d x

\frac{8 x ^{3}}{3} + \frac{3 x ^{5}}{5} - \frac{5}{14 x ^{7}} + e^{x} + C

Solution steps

\int 8 x^{2} + 3 x^{4} + \frac{5}{2 x ^{8}} + e^{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 8 x^{2} d x + \int 3 x^{4} d x + \int \frac{5}{2 x ^{8}} d x + \int e^{x} d x

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

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

\int \frac{5}{2 x ^{8}} d x = - \frac{5}{14 x ^{7}}

\int e^{x} d x = e^{x}

= \frac{8 x ^{3}}{3} + \frac{3 x ^{5}}{5} - \frac{5}{14 x ^{7}} + e^{x}

Add a constant to the solution

= \frac{8 x ^{3}}{3} + \frac{3 x ^{5}}{5} - \frac{5}{14 x ^{7}} + e^{x} + C

x y^{^{'}} - y = 2020 x^{2}, y (1) = 1

\int 12 sin^{3} (x co) s^{2} x d x

y^{^{'}} - \frac{1}{x} y = 9, y (1) = \frac{1}{7}

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

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

What is the integral of (8x^2+3x^4+5/(2x^8)+e^x) ?
The integral of (8x^2+3x^4+5/(2x^8)+e^x) is (8x^3)/3+(3x^5)/5-5/(14x^7)+e^x+C