What is the integral of e^xsqrt(35+e^x) ?

The integral of e^xsqrt(35+e^x) is 2/3 (35+e^x)^{3/2}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of e^xsqrt(35+e^x)

\int e^{x} 35 + e^{x} d x

\frac{2}{3} (35 + e^{x})^{\frac{3}{2}} + C

Solution steps

\int e^{x} 35 + e^{x} d x

Apply u-substitution

: \int u d u

= \int u d u

Apply the Power Rule

: \frac{2}{3} u^{\frac{3}{2}}

= \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 35 + e^{x}

= \frac{2}{3} (35 + e^{x})^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{3} (35 + e^{x})^{\frac{3}{2}} + C

\int \frac{6}{x ^{2} + 1} - x^{2} d x

\frac{\partial}{\partial x} (ln (1 + \frac{1}{x}))

\frac{\partial}{\partial x} (x^{2} + 2 x y + 3 y^{2})

\int \frac{1}{u} d u

y^{^{''}} - 2 y^{^{'}} - 15 y = 0

What is the integral of e^xsqrt(35+e^x) ?
The integral of e^xsqrt(35+e^x) is 2/3 (35+e^x)^{3/2}+C