What is the integral of 1/(4u^2+23) ?

The integral of 1/(4u^2+23) is 1/(2sqrt(23))arctan(2/(sqrt(23))u)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(4u^2+23)

\int \frac{1}{4 u ^{2} + 23} d u

\frac{1}{2 23} arctan (\frac{2}{23} u) + C

Solution steps

\int \frac{1}{4 u ^{2} + 23} d u

Apply Integral Substitution

= \int \frac{1}{2 23 ( v ^{2} + 1 )} d v

Take the constant out:

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

= \frac{1}{2 23} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

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

= \frac{1}{2 23} arctan (v)

Substitute back

v = \frac{2}{23} u

= \frac{1}{2 23} arctan (\frac{2}{23} u)

Add a constant to the solution

= \frac{1}{2 23} arctan (\frac{2}{23} u) + C

\frac{d}{d x} ((2 + x) x)

f (x) = x^{2} + 9, (- 5, 34)

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

1 - x^{2}, 4 - 4 x

What is the integral of 1/(4u^2+23) ?
The integral of 1/(4u^2+23) is 1/(2sqrt(23))arctan(2/(sqrt(23))u)+C