What is the integral of (-5)/(x^2sqrt(5+x^2)) ?

The integral of (-5)/(x^2sqrt(5+x^2)) is (sqrt(5+x^2))/x+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (-5)/(x^2sqrt(5+x^2))

\int \frac{- 5}{x ^{2} 5 + x ^{2}} d x

\frac{5 + x ^{2}}{x} + C

Solution steps

\int \frac{- 5}{x ^{2} 5 + x ^{2}} d x

Take the constant out:

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

= - 5 \cdot \int \frac{1}{x ^{2} 5 + x ^{2}} d x

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{5 tan ^{2} ( u )} d u

= - 5 \cdot \int \frac{sec ( u )}{5 tan ^{2} ( u )} d u

Take the constant out:

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

= - 5 \cdot \frac{1}{5} \cdot \int \frac{sec ( u )}{tan ^{2} ( u )} d u

Rewrite using trig identities

= - 5 \cdot \frac{1}{5} \cdot \int \frac{cos ( u )}{sin ^{2} ( u )} d u

Apply u-substitution

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

= - 5 \cdot \frac{1}{5} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= - 5 \cdot \frac{1}{5} (- \frac{1}{v})

Substitute back

= - 5 \cdot \frac{1}{5} - \frac{1}{sin ( arctan ( \frac{1}{5} x ) )}

Simplify

- 5 \cdot \frac{1}{5} - \frac{1}{sin ( arctan ( \frac{1}{5} x ) )} : \frac{5 + x ^{2}}{x}

= \frac{5 + x ^{2}}{x}

Add a constant to the solution

= \frac{5 + x ^{2}}{x} + C

\int \frac{cos ( \frac{π}{x ^{4}} )}{x ^{5}} d x

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

\int \frac{1}{( arctan ( y ) ) ( 1 + y ^{2} )} d y

\int (3 - cos (x))^{2} d x

\int \frac{1}{x} ln^{2} (x) d x

What is the integral of (-5)/(x^2sqrt(5+x^2)) ?
The integral of (-5)/(x^2sqrt(5+x^2)) is (sqrt(5+x^2))/x+C