What is the integral of (1+3x^2)/(x^2(1+x^2)) ?

The integral of (1+3x^2)/(x^2(1+x^2)) is -1/x+2arctan(x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (1+3x^2)/(x^2(1+x^2))

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

- \frac{1}{x} + 2 arctan (x) + C

Solution steps

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

Expand

\frac{1 + 3 x ^{2}}{x ^{2} ( 1 + x ^{2} )} : \frac{1}{x ^{2} ( 1 + x ^{2} )} + \frac{3 x ^{2}}{x ^{2} ( 1 + x ^{2} )}

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

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

\int \frac{3 x ^{2}}{x ^{2} ( 1 + x ^{2} )} d x = 3 arctan (x)

= - \frac{1}{x} - arctan (x) + 3 arctan (x)

Simplify

= - \frac{1}{x} + 2 arctan (x)

Add a constant to the solution

= - \frac{1}{x} + 2 arctan (x) + C

\int c^{a x} d x

\int \frac{5}{x \cdot x ^{5} + 4} d x

\int \frac{8 x + 6}{( 4 x ^{2} + 6 x ) ^{3}} d x

\int x^{4} (x^{5} - 7)^{4} d x

\int ∣ x + 3 ∣ d x

What is the integral of (1+3x^2)/(x^2(1+x^2)) ?
The integral of (1+3x^2)/(x^2(1+x^2)) is -1/x+2arctan(x)+C