What is the integral of p/((p^2+4)) ?

The integral of p/((p^2+4)) is 1/2 ln|p^2+4|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of p/((p^2+4))

integral

\frac{p}{( p ^{2} + 4 )}

\frac{1}{2} ln p^{2} + 4 + C

Solution steps

\int \frac{p}{p ^{2} + 4} d p

Apply u-substitution

: \int \frac{1}{2 u} d u

= \int \frac{1}{2 u} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} ln ∣ u ∣

Substitute back

u = p^{2} + 4

= \frac{1}{2} ln p^{2} + 4

Add a constant to the solution

= \frac{1}{2} ln p^{2} + 4 + C

4 x^{2} + 3 x + 2 \cdot 2 (x + 1)

(8 x - x^{4})^{3} (8 x^{3} - 16)

cos^{2} (\frac{x}{4})

(\frac{1}{x ^{4}}) sin (x)

cos ((\frac{1}{4}) x)

What is the integral of p/((p^2+4)) ?
The integral of p/((p^2+4)) is 1/2 ln|p^2+4|+C