What is the integral of 5/(x^2+3) ?

The integral of 5/(x^2+3) is 5/(sqrt(3))arctan(x/(sqrt(3)))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 5/(x^2+3)

\int \frac{5}{x ^{2} + 3} d x

\frac{5}{3} arctan (\frac{x}{3}) + C

Solution steps

\int \frac{5}{x ^{2} + 3} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 5 \cdot \int \frac{1}{x ^{2} + 3} d x

Apply Integral Substitution

= 5 \cdot \int \frac{1}{3 ( u ^{2} + 1 )} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 5 \cdot \frac{1}{3} \cdot \int \frac{1}{u ^{2} + 1} d u

Use the common integral:

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

= 5 \cdot \frac{1}{3} arctan (u)

Substitute back

u = \frac{x}{3}

= 5 \cdot \frac{1}{3} arctan (\frac{x}{3})

Simplify

5 \cdot \frac{1}{3} arctan (\frac{x}{3}) : \frac{5}{3} arctan (\frac{x}{3})

= \frac{5}{3} arctan (\frac{x}{3})

Add a constant to the solution

= \frac{5}{3} arctan (\frac{x}{3}) + C

\int \frac{1}{1 + 4 x - x ^{2}} d x

\int sec (\frac{1}{2}) y d y

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

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

\int 9 cos^{2} (3 x) d x

What is the integral of 5/(x^2+3) ?
The integral of 5/(x^2+3) is 5/(sqrt(3))arctan(x/(sqrt(3)))+C