What is the integral of e^xarccos(e^x) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of e^xarccos(e^x)

\int e^{x} arccos (e^{x}) d x

e^{x} arccos (e^{x}) - 1 - e^{2 x} + C

Solution steps

\int e^{x} arccos (e^{x}) d x

Apply Integration By Parts

= e^{x} arccos (e^{x}) - \int - \frac{e ^{2 x}}{1 - e ^{2 x}} d x

\int - \frac{e ^{2 x}}{1 - e ^{2 x}} d x = 1 - e^{2 x}

= e^{x} arccos (e^{x}) - 1 - e^{2 x}

Add a constant to the solution

= e^{x} arccos (e^{x}) - 1 - e^{2 x} + C

\int - x^{2} - 6 x - 5 d x

\int (12 y cos (4 x) - 3 sin (y)) d x

\int 6 x e^{x^{2} + 3} d x

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

\int \frac{1}{x ( 1 + x ^{6} )} d x

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