What is the integral of 1/(x^2sqrt(4+x^2)) ?

The integral of 1/(x^2sqrt(4+x^2)) is -(sqrt(4+x^2))/(4x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(x^2sqrt(4+x^2))

\int \frac{1}{x ^{2} 4 + x ^{2}} d x

- \frac{4 + x ^{2}}{4 x} + C

Solution steps

\int \frac{1}{x ^{2} 4 + x ^{2}} d x

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{4 tan ^{2} ( u )} d u

= \int \frac{sec ( u )}{4 tan ^{2} ( u )} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{sec ( u )}{tan ^{2} ( u )} d u

Rewrite using trig identities

= \frac{1}{4} \cdot \int \frac{cos ( u )}{sin ^{2} ( u )} d u

Apply u-substitution

: \int \frac{1}{v ^{2}} d v

= \frac{1}{4} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= \frac{1}{4} (- \frac{1}{v})

Substitute back

= \frac{1}{4} (- \frac{1}{sin ( arctan ( \frac{1}{2} x ) )})

Simplify

\frac{1}{4} (- \frac{1}{sin ( arctan ( \frac{1}{2} x ) )}) : - \frac{4 + x ^{2}}{4 x}

= - \frac{4 + x ^{2}}{4 x}

Add a constant to the solution

= - \frac{4 + x ^{2}}{4 x} + C

\int \frac{e ^{x}}{e ^{2 x} - 3 e ^{x} + 2} d x

\frac{d}{d x} (tan^{2} (3 x) + 1)

\int t^{2} e^{- \frac{t}{3}} d t

n = 7 \sum \infty \frac{1}{6 n}

y = x^{2} + 5 x, x = 8

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