What is the integral of e^xsqrt(e^{2x)}+6e^x+10 ?

The integral of e^xsqrt(e^{2x)}+6e^x+10 is 1/2 e^{2x}+6e^x+10x+C

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integral of e^xsqrt(e^{2x)}+6e^x+10

integral

e^{x} e^{2 x} + 6 e^{x} + 10

\frac{1}{2} e^{2 x} + 6 e^{x} + 10 x + C

Solution steps

\int e^{x} e^{2 x} + 6 e^{x} + 10 d x

e^{x} e^{2 x} = e^{2 x}

= \int e^{2 x} + 6 e^{x} + 10 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 e^{2 x} d x + \int 6 e^{x} d x + \int 10 d x

\int e^{2 x} d x = \frac{1}{2} e^{2 x}

\int 6 e^{x} d x = 6 e^{x}

\int 10 d x = 10 x

= \frac{1}{2} e^{2 x} + 6 e^{x} + 10 x

Add a constant to the solution

= \frac{1}{2} e^{2 x} + 6 e^{x} + 10 x + C

1 - cos (2 x)

2 e^{2} x

lo g_{10} (\frac{2}{y})

\frac{1}{( x ^{5} + x ^{3} )}

t + 2

What is the integral of e^xsqrt(e^{2x)}+6e^x+10 ?
The integral of e^xsqrt(e^{2x)}+6e^x+10 is 1/2 e^{2x}+6e^x+10x+C