What is the integral of (-2x^5-5x^3+3x^2+1-4e^x) ?

The integral of (-2x^5-5x^3+3x^2+1-4e^x) is -(x^6)/3-(5x^4)/4+x^3+x-4e^x+C

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integral of (-2x^5-5x^3+3x^2+1-4e^x)

\int (- 2 x^{5} - 5 x^{3} + 3 x^{2} + 1 - 4 e^{x}) d x

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

Solution steps

\int - 2 x^{5} - 5 x^{3} + 3 x^{2} + 1 - 4 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 2 x^{5} d x - \int 5 x^{3} d x + \int 3 x^{2} d x + \int 1 d x - \int 4 e^{x} d x

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

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

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

\int 1 d x = x

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

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

Add a constant to the solution

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

\frac{d y}{d x} + (\frac{4}{76 + 3 x}) y = 4.2

\frac{d}{d x} (7 x - 2)

\frac{d}{d t} (4 sin (t))

y^{^{''}} + 11 y^{^{'}} + 18 y = 0, y (0) = 1, y^{^{'}} (0) = 0

What is the integral of (-2x^5-5x^3+3x^2+1-4e^x) ?
The integral of (-2x^5-5x^3+3x^2+1-4e^x) is -(x^6)/3-(5x^4)/4+x^3+x-4e^x+C