What is the integral of (-7)/(sqrt(x^2+16)) ?

The integral of (-7)/(sqrt(x^2+16)) is -7ln(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 (-7)/(sqrt(x^2+16))

\int \frac{- 7}{x ^{2} + 16} d x

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

Solution steps

\int \frac{- 7}{x ^{2} + 16} d x

Take the constant out:

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

= - 7 \cdot \int \frac{1}{x ^{2} + 16} d x

Apply Trigonometric Substitution

: \int sec (u) d u

= - 7 \cdot \int sec (u) d u

Use the common integral:

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

= - 7 ln ∣ tan (u) + sec (u) ∣

Substitute back

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

= - 7 ln tan (arctan (\frac{1}{4} x)) + sec (arctan (\frac{1}{4} x))

Simplify

- 7 ln tan (arctan (\frac{1}{4} x)) + sec (arctan (\frac{1}{4} x)) : - 7 ln (\frac{1}{4} x + 16 + x^{2})

= - 7 ln (\frac{1}{4} x + 16 + x^{2})

Add a constant to the solution

= - 7 ln (\frac{1}{4} x + 16 + x^{2}) + C

x \to 5 lim (\frac{( ln ( \frac{x}{5} ) )}{x ^{2} - 25})

\int \frac{1}{ln ( x ^{x} )} d x

\int \frac{x}{12} d x

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

\frac{sin ( x ^{2} )}{x ^{2}}

What is the integral of (-7)/(sqrt(x^2+16)) ?
The integral of (-7)/(sqrt(x^2+16)) is -7ln(1/4 |x+sqrt(16+x^2)|)+C