What is the integral of (((7-2x))/(x^{1/2)}+3(5-e^x+2e^3)) ?

The integral of (((7-2x))/(x^{1/2)}+3(5-e^x+2e^3)) is 14x^{1/2}-4/3 x^{3/2}+3(5x-e^x+2e^3x)+C

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integral of (((7-2x))/(x^{1/2)}+3(5-e^x+2e^3))

\int (\frac{( 7 - 2 x )}{x ^{\frac{1}{2}}} + 3 (5 - e^{x} + 2 e^{3})) d x

14 x^{\frac{1}{2}} - \frac{4}{3} x^{\frac{3}{2}} + 3 (5 x - e^{x} + 2 e^{3} x) + C

Solution steps

\int \frac{7 - 2 x}{x ^{\frac{1}{2}}} + 3 (5 - e^{x} + 2 e^{3}) 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 \frac{7 - 2 x}{x ^{\frac{1}{2}}} d x + \int 3 (5 - e^{x} + 2 e^{3}) d x

\int \frac{7 - 2 x}{x ^{\frac{1}{2}}} d x = 14 x^{\frac{1}{2}} - \frac{4}{3} x^{\frac{3}{2}}

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

= 14 x^{\frac{1}{2}} - \frac{4}{3} x^{\frac{3}{2}} + 3 (5 x - e^{x} + 2 e^{3} x)

Add a constant to the solution

= 14 x^{\frac{1}{2}} - \frac{4}{3} x^{\frac{3}{2}} + 3 (5 x - e^{x} + 2 e^{3} x) + C

x^{3} e^{x}

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\frac{d y}{d x} = \frac{1}{x ^{2} + 1}

What is the integral of (((7-2x))/(x^{1/2)}+3(5-e^x+2e^3)) ?
The integral of (((7-2x))/(x^{1/2)}+3(5-e^x+2e^3)) is 14x^{1/2}-4/3 x^{3/2}+3(5x-e^x+2e^3x)+C