What is the integral of (e^x2^{e^x}+2xe^{x^2}) ?

The integral of (e^x2^{e^x}+2xe^{x^2}) is (2^{e^x})/(ln(2))+e^{x^2}+C

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integral of (e^x2^{e^x}+2xe^{x^2})

\int (e^{x} 2^{e^{x}} + 2 x e^{x^{2}}) d x

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

Solution steps

\int e^{x} \cdot 2^{e^{x}} + 2 x e^{x^{2}} d x

Apply u-substitution

: \int \frac{2 ^{u} u + 2 e ^{l n^{2} (u)} ln ( u )}{u} d u

= \int \frac{2 ^{u} u + 2 e ^{l n^{2} (u)} ln ( u )}{u} d u

Expand

\frac{2 ^{u} u + 2 e ^{l n^{2} (u)} ln ( u )}{u} : 2^{u} + \frac{2 e ^{l n^{2} (u)} ln ( u )}{u}

= \int 2^{u} + \frac{2 e ^{l n^{2} (u)} ln ( u )}{u} d u

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^{u} d u + \int \frac{2 e ^{l n^{2} (u)} ln ( u )}{u} d u

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

\int \frac{2 e ^{l n^{2} (u)} ln ( u )}{u} d u = e^{l n^{2} (u)}

= \frac{2 ^{u}}{ln ( 2 )} + e^{l n^{2} (u)}

Substitute back

u = e^{x}

= \frac{2 ^{e^{x}}}{ln ( 2 )} + e^{l n^{2} (e^{x})}

Simplify

ln (e^{x}) : x

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

Add a constant to the solution

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

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What is the integral of (e^x2^{e^x}+2xe^{x^2}) ?
The integral of (e^x2^{e^x}+2xe^{x^2}) is (2^{e^x})/(ln(2))+e^{x^2}+C