What is the integral of sqrt(x)-2^x-3sin(3x)+4e^{2x} ?

The integral of sqrt(x)-2^x-3sin(3x)+4e^{2x} is 2/3 x^{3/2}-(2^x)/(ln(2))+cos(3x)+2e^{2x}+C

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integral of sqrt(x)-2^x-3sin(3x)+4e^{2x}

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

\frac{2}{3} x^{\frac{3}{2}} - \frac{2 ^{x}}{ln ( 2 )} + cos (3 x) + 2 e^{2 x} + C

Solution steps

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

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

\int 2^{x} d x = \frac{2 ^{x}}{ln ( 2 )}

\int 3 sin (3 x) d x = - cos (3 x)

\int 4 e^{2 x} d x = 2 e^{2 x}

= \frac{2}{3} x^{\frac{3}{2}} - \frac{2 ^{x}}{ln ( 2 )} - (- cos (3 x)) + 2 e^{2 x}

Simplify

= \frac{2}{3} x^{\frac{3}{2}} - \frac{2 ^{x}}{ln ( 2 )} + cos (3 x) + 2 e^{2 x}

Add a constant to the solution

= \frac{2}{3} x^{\frac{3}{2}} - \frac{2 ^{x}}{ln ( 2 )} + cos (3 x) + 2 e^{2 x} + C

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

\frac{d y}{d x} - 2 x y = 2 x

x y^{^{'}} = y + 5 x^{2} sin (x), y (π) = 0

\frac{\partial}{\partial x} (y x^{2})

\int \frac{8}{x + 8 + x} d x

What is the integral of sqrt(x)-2^x-3sin(3x)+4e^{2x} ?
The integral of sqrt(x)-2^x-3sin(3x)+4e^{2x} is 2/3 x^{3/2}-(2^x)/(ln(2))+cos(3x)+2e^{2x}+C