What is the integral of 4/(sqrt(x^2+9)) ?

The integral of 4/(sqrt(x^2+9)) is 4ln(1/3 |x+sqrt(9+x^2)|)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 4/(sqrt(x^2+9))

\int \frac{4}{x ^{2} + 9} d x

4 ln (\frac{1}{3} x + 9 + x^{2}) + C

Solution steps

\int \frac{4}{x ^{2} + 9} d x

Take the constant out:

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

= 4 \cdot \int \frac{1}{x ^{2} + 9} d x

Apply Trigonometric Substitution

: \int sec (u) d u

= 4 \cdot \int sec (u) d u

Use the common integral:

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

= 4 ln ∣ tan (u) + sec (u) ∣

Substitute back

u = arctan (\frac{1}{3} x)

= 4 ln tan (arctan (\frac{1}{3} x)) + sec (arctan (\frac{1}{3} x))

Simplify

4 ln tan (arctan (\frac{1}{3} x)) + sec (arctan (\frac{1}{3} x)) : 4 ln (\frac{1}{3} x + 9 + x^{2})

= 4 ln (\frac{1}{3} x + 9 + x^{2})

Add a constant to the solution

= 4 ln (\frac{1}{3} x + 9 + x^{2}) + C

6 x y

\int x x - 30 d x

\int_{0}^{\infty} 8 e^{- 4 x} d x

\int_{0}^{5} \frac{1}{4} (5 - x) x d x

\frac{d}{d t} (v (t) t)

What is the integral of 4/(sqrt(x^2+9)) ?
The integral of 4/(sqrt(x^2+9)) is 4ln(1/3 |x+sqrt(9+x^2)|)+C