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

The integral of ((2x)/(x^2+1)+sec^2(x)) is ln|x^2+1|+tan(x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of ((2x)/(x^2+1)+sec^2(x))

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

ln x^{2} + 1 + tan (x) + C

Solution steps

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

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

\int sec^{2} (x) d x = tan (x)

= ln x^{2} + 1 + tan (x)

Add a constant to the solution

= ln x^{2} + 1 + tan (x) + C

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

\int (3 x^{2} + 5 x - 4) d x

\int \frac{11}{3 - 2 x} d x

\int 3 + e^{x} d x

\int lo g_{e^{1}} (x) d x

What is the integral of ((2x)/(x^2+1)+sec^2(x)) ?
The integral of ((2x)/(x^2+1)+sec^2(x)) is ln|x^2+1|+tan(x)+C