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

The integral of (e^x)/(sqrt(-e^{2x)-4e^x)} is arcsin(1/2 e^x+1)+C

integral of (e^x)/(sqrt(-e^{2x)-4e^x)}

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

arcsin (\frac{1}{2} e^{x} + 1) + C

Solution steps

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

Apply u-substitution

= \int \frac{1}{- u ^{2} - 4 u} d u

Complete the square

- u^{2} - 4 u : - (u + 2)^{2} + 4

= \int \frac{1}{- ( u + 2 ) ^{2} + 4} d u

Apply u-substitution

= \int \frac{1}{- v ^{2} + 4} d v

Apply Trigonometric Substitution

= \int 1 d w

Integral of a constant:

\int a d x = a x

= 1 \cdot w

Substitute back

= 1 \cdot arcsin (\frac{1}{2} (e^{x} + 2))

Simplify

1 \cdot arcsin (\frac{1}{2} (e^{x} + 2)) : arcsin (\frac{1}{2} e^{x} + 1)

= arcsin (\frac{1}{2} e^{x} + 1)

Add a constant to the solution

= arcsin (\frac{1}{2} e^{x} + 1) + C

\int (tan (5 w) + cot (5 w))^{2} d w

\int \frac{x ^{4} + 3 x ^{3} + x + 1}{x ^{2} + 1} d x

\int 31 tan^{5} (x) sec^{4} (x) d x

\int 5 x^{4} + 2 x^{- 3} - 3 e^{x} + \frac{2}{3} x^{- 1} d x

\int \frac{x ^{2} - 3}{x ^{2} + 1} d x

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