What is the integral of 1/(sqrt(2^2+u^2)) ?

The integral of 1/(sqrt(2^2+u^2)) is ln(1/2 |u+sqrt(4+u^2)|)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(sqrt(2^2+u^2))

\int \frac{1}{2 ^{2} + u ^{2}} d u

ln (\frac{1}{2} u + 4 + u^{2}) + C

Solution steps

\int \frac{1}{2 ^{2} + u ^{2}} d u

Simplify

= \int \frac{1}{4 + u ^{2}} d u

Apply Trigonometric Substitution

: \int sec (v) d v

= \int sec (v) d v

Use the common integral:

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

= ln ∣ tan (v) + sec (v) ∣

Substitute back

v = arctan (\frac{1}{2} u)

= ln tan (arctan (\frac{1}{2} u)) + sec (arctan (\frac{1}{2} u))

Simplify

ln tan (arctan (\frac{1}{2} u)) + sec (arctan (\frac{1}{2} u)) : ln (\frac{1}{2} u + 4 + u^{2})

= ln (\frac{1}{2} u + 4 + u^{2})

Add a constant to the solution

= ln (\frac{1}{2} u + 4 + u^{2}) + C

\int \frac{1}{- x ^{2} - 14 x - 48} d x

\int_{0}^{4} 8 x d x

\int \frac{x + 1}{x ^{2} + 2 x} d x

\frac{d y}{d x} = cos (2 x)

y^{^{''}} - 7 y^{^{'}} + 12 y = 6 e^{t}

What is the integral of 1/(sqrt(2^2+u^2)) ?
The integral of 1/(sqrt(2^2+u^2)) is ln(1/2 |u+sqrt(4+u^2)|)+C