What is the integral of-(10)/(sqrt(x^2+16)) ?

The integral of-(10)/(sqrt(x^2+16)) is -10ln(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-(10)/(sqrt(x^2+16))

\int - \frac{10}{x ^{2} + 16} d x

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

Solution steps

\int - \frac{10}{x ^{2} + 16} d x

Take the constant out:

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

= - 10 \cdot \int \frac{1}{x ^{2} + 16} d x

Apply Trigonometric Substitution

: \int sec (u) d u

= - 10 \cdot \int sec (u) d u

Use the common integral:

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

= - 10 ln ∣ tan (u) + sec (u) ∣

Substitute back

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

= - 10 ln tan (arctan (\frac{1}{4} x)) + sec (arctan (\frac{1}{4} x))

Simplify

- 10 ln tan (arctan (\frac{1}{4} x)) + sec (arctan (\frac{1}{4} x)) : - 10 ln (\frac{1}{4} x + 16 + x^{2})

= - 10 ln (\frac{1}{4} x + 16 + x^{2})

Add a constant to the solution

= - 10 ln (\frac{1}{4} x + 16 + x^{2}) + C

\int f (u) d u

f (x) = \frac{5 x}{x - 2}, (3, 15)

- y^{^{''}} + 36 y = 0

\frac{d}{d x} (x^{\frac{7}{2}})

6 y

What is the integral of-(10)/(sqrt(x^2+16)) ?
The integral of-(10)/(sqrt(x^2+16)) is -10ln(1/4 |x+sqrt(16+x^2)|)+C