What is the integral of e^x-1/(1+x^2) ?

The integral of e^x-1/(1+x^2) is e^x-arctan(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-1/(1+x^2)

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

e^{x} - arctan (x) + C

Solution steps

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int e^{x} d x = e^{x}

\int \frac{1}{1 + x ^{2}} d x = arctan (x)

= e^{x} - arctan (x)

Add a constant to the solution

= e^{x} - arctan (x) + C

\frac{d}{d x} (tan (x^{2} - 9))

\int cos (ln (x^{2})) d x

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

f (x) = \frac{6}{x}, at x = 4

3 y^{^{'}} + 5 y = 1

What is the integral of e^x-1/(1+x^2) ?
The integral of e^x-1/(1+x^2) is e^x-arctan(x)+C