What is the integral of (200)/(sqrt(16x^2+256)) ?

The integral of (200)/(sqrt(16x^2+256)) is 50ln(1/4 |x+sqrt(x^2+16)|)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (200)/(sqrt(16x^2+256))

\int \frac{200}{16 x ^{2} + 256} d x

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

Solution steps

\int \frac{200}{16 x ^{2} + 256} d x

Take the constant out:

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

= 200 \cdot \int \frac{1}{16 x ^{2} + 256} d x

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{4} d u

= 200 \cdot \int \frac{sec ( u )}{4} d u

Take the constant out:

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

= 200 \cdot \frac{1}{4} \cdot \int sec (u) d u

Use the common integral:

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

= 200 \cdot \frac{1}{4} ln ∣ tan (u) + sec (u) ∣

Substitute back

u = arctan (\frac{4}{16} x)

= 200 \cdot \frac{1}{4} ln tan (arctan (\frac{4}{16} x)) + sec (arctan (\frac{4}{16} x))

Simplify

200 \cdot \frac{1}{4} ln tan (arctan (\frac{4}{16} x)) + sec (arctan (\frac{4}{16} x)) : 50 ln (\frac{1}{4} x + x^{2} + 16)

= 50 ln (\frac{1}{4} x + x^{2} + 16)

Add a constant to the solution

= 50 ln (\frac{1}{4} x + x^{2} + 16) + C

\frac{d ^{2} y}{d x ^{2}} = 4 y, y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} - y^{^{'}} = x e^{x}

12 = \frac{1}{( 22 \cdot 1 0 ^{- 6} )} y + \frac{33000 d y}{d x}

\frac{\partial}{\partial x} (ln (y) + 4 x)

x \to - 3 lim (\frac{16 - x ^{\frac{4}{3}}}{4 - x ^{\frac{2}{3}}})

What is the integral of (200)/(sqrt(16x^2+256)) ?
The integral of (200)/(sqrt(16x^2+256)) is 50ln(1/4 |x+sqrt(x^2+16)|)+C