What is the integral of sqrt(4+e^x) ?

The integral of sqrt(4+e^x) is -2ln| 1/2 sqrt(e^x+4)+1|+2ln| 1/2 sqrt(e^x+4)-1|+2sqrt(e^x+4)+C

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integral of sqrt(4+e^x)

\int 4 + e^{x} d x

- 2 ln \frac{1}{2} e^{x} + 4 + 1 + 2 ln \frac{1}{2} e^{x} + 4 - 1 + 2 e^{x} + 4 + C

Solution steps

\int 4 + e^{x} d x

Apply Integration By Parts

: x 4 + e^{x} - \int \frac{e ^{x} x}{2 4 + e ^{x}} d x

= x 4 + e^{x} - \int \frac{e ^{x} x}{2 4 + e ^{x}} d x

\int \frac{e ^{x} x}{2 4 + e ^{x}} d x = x e^{x} + 4 + 2 (ln \frac{1}{2} e^{x} + 4 + 1 - ln \frac{1}{2} e^{x} + 4 - 1 - e^{x} + 4)

= x 4 + e^{x} - (x e^{x} + 4 + 2 (ln \frac{1}{2} e^{x} + 4 + 1 - ln \frac{1}{2} e^{x} + 4 - 1 - e^{x} + 4))

Simplify

x 4 + e^{x} - (x e^{x} + 4 + 2 (ln \frac{1}{2} e^{x} + 4 + 1 - ln \frac{1}{2} e^{x} + 4 - 1 - e^{x} + 4)) : - 2 ln \frac{1}{2} e^{x} + 4 + 1 + 2 ln \frac{1}{2} e^{x} + 4 - 1 + 2 e^{x} + 4

= - 2 ln \frac{1}{2} e^{x} + 4 + 1 + 2 ln \frac{1}{2} e^{x} + 4 - 1 + 2 e^{x} + 4

Add a constant to the solution

= - 2 ln \frac{1}{2} e^{x} + 4 + 1 + 2 ln \frac{1}{2} e^{x} + 4 - 1 + 2 e^{x} + 4 + C

\frac{d}{d x} (\frac{1}{16} x^{4})

\int 6 sec^{4} (x) d x

\int_{0}^{1} 2 x - 2 x^{2} d x

f (x) = \frac{3}{1 + x ^{2}}

(7^{x})^{^{'}}

What is the integral of sqrt(4+e^x) ?
The integral of sqrt(4+e^x) is -2ln| 1/2 sqrt(e^x+4)+1|+2ln| 1/2 sqrt(e^x+4)-1|+2sqrt(e^x+4)+C