What is the integral of (100)/(sqrt(36+25x^2)) ?

The integral of (100)/(sqrt(36+25x^2)) is 20ln(1/6 |5x+sqrt(25x^2+36)|)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (100)/(sqrt(36+25x^2))

\int \frac{100}{36 + 25 x ^{2}} d x

20 ln (\frac{1}{6} 5 x + 25 x^{2} + 36) + C

Solution steps

\int \frac{100}{36 + 25 x ^{2}} d x

Take the constant out:

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

= 100 \cdot \int \frac{1}{36 + 25 x ^{2}} d x

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{5} d u

= 100 \cdot \int \frac{sec ( u )}{5} d u

Take the constant out:

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

= 100 \cdot \frac{1}{5} \cdot \int sec (u) d u

Use the common integral:

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

= 100 \cdot \frac{1}{5} ln ∣ tan (u) + sec (u) ∣

Substitute back

u = arctan (\frac{5}{6} x)

= 100 \cdot \frac{1}{5} ln tan (arctan (\frac{5}{6} x)) + sec (arctan (\frac{5}{6} x))

Simplify

100 \cdot \frac{1}{5} ln tan (arctan (\frac{5}{6} x)) + sec (arctan (\frac{5}{6} x)) : 20 ln (\frac{1}{6} 5 x + 25 x^{2} + 36)

= 20 ln (\frac{1}{6} 5 x + 25 x^{2} + 36)

Add a constant to the solution

= 20 ln (\frac{1}{6} 5 x + 25 x^{2} + 36) + C

n = 1 \sum \infty n e^{- n}

2 y^{^{'}} = y^{3} \cdot cos (x)

\int e^{4 x - 5} d x

\int \frac{x ^{2} - 8}{4 x - x ^{3}} d x

- \frac{x}{6 y}

What is the integral of (100)/(sqrt(36+25x^2)) ?
The integral of (100)/(sqrt(36+25x^2)) is 20ln(1/6 |5x+sqrt(25x^2+36)|)+C