What is the integral of (e^x)/(sqrt(4-9e^{2x))} ?

The integral of (e^x)/(sqrt(4-9e^{2x))} is 1/3 arcsin(3/2 e^x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (e^x)/(sqrt(4-9e^{2x))}

\int \frac{e ^{x}}{4 - 9 e ^{2 x}} d x

\frac{1}{3} arcsin (\frac{3}{2} e^{x}) + C

Solution steps

\int \frac{e ^{x}}{4 - 9 e ^{2 x}} d x

Apply u-substitution

: \int \frac{1}{4 - 9 u ^{2}} d u

= \int \frac{1}{4 - 9 u ^{2}} d u

Apply Trigonometric Substitution

: \int \frac{1}{3} d v

= \int \frac{1}{3} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= \frac{1}{3} \cdot 1 \cdot v

Substitute back

= \frac{1}{3} \cdot 1 \cdot arcsin (\frac{3}{2} e^{x})

Simplify

= \frac{1}{3} arcsin (\frac{3}{2} e^{x})

Add a constant to the solution

= \frac{1}{3} arcsin (\frac{3}{2} e^{x}) + C

\frac{\partial}{\partial x} (5 - x^{4} + 2 x^{2} - y^{2})

(cos (9 x))^{x}

48 y^{^{''}} + 34 y^{^{'}} - 5 y = 0

\frac{d}{d x} ((4 x - 3)^{2})

x^{(\frac{4}{3})}, 2 x^{(\frac{1}{3})}, 0, 2

What is the integral of (e^x)/(sqrt(4-9e^{2x))} ?
The integral of (e^x)/(sqrt(4-9e^{2x))} is 1/3 arcsin(3/2 e^x)+C