What is the integral of (-6)/(sqrt(x^2+16)) ?

The integral of (-6)/(sqrt(x^2+16)) is -6ln(1/4 |x+sqrt(16+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 (-6)/(sqrt(x^2+16))

\int \frac{- 6}{x ^{2} + 16} d x

- 6 ln (\frac{1}{4} x + 16 + x^{2}) + C

Solution steps

\int \frac{- 6}{x ^{2} + 16} d x

Take the constant out:

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

= - 6 \cdot \int \frac{1}{x ^{2} + 16} d x

Apply Trigonometric Substitution

: \int sec (u) d u

= - 6 \cdot \int sec (u) d u

Use the common integral:

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

= - 6 ln ∣ tan (u) + sec (u) ∣

Substitute back

u = arctan (\frac{1}{4} x)

= - 6 ln tan (arctan (\frac{1}{4} x)) + sec (arctan (\frac{1}{4} x))

Simplify

- 6 ln tan (arctan (\frac{1}{4} x)) + sec (arctan (\frac{1}{4} x)) : - 6 ln (\frac{1}{4} x + 16 + x^{2})

= - 6 ln (\frac{1}{4} x + 16 + x^{2})

Add a constant to the solution

= - 6 ln (\frac{1}{4} x + 16 + x^{2}) + C

x \to 1 + lim (\frac{2}{1 - x})

\frac{d}{d x} (a - b)

g (t) = t^{2} - \frac{2}{t ^{3}}

\int t^{2} e^{- 20 t} d t

\int_{t}^{1} ln (x) d x

What is the integral of (-6)/(sqrt(x^2+16)) ?
The integral of (-6)/(sqrt(x^2+16)) is -6ln(1/4 |x+sqrt(16+x^2)|)+C