What is the integral of 1/(x^2*sqrt(5+x^2)) ?

The integral of 1/(x^2*sqrt(5+x^2)) is -(sqrt(5+x^2))/(5x)+C

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integral of 1/(x^2*sqrt(5+x^2))

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

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

Solution steps

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

Apply Trigonometric Substitution

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

= \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

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

Rewrite using trig identities

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

Apply u-substitution

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

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

Apply the Power Rule

: - \frac{1}{v}

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

y^{^{''}} - 8 y^{^{'}} + 18 y = 0, y (0) = - 4, y^{^{'}} (0) = 3

\int \frac{( 1 - cos ( 2 x ) )}{2} d x

\frac{sin ( θ )}{7} + \frac{c}{θ}

x \to 3 lim (\frac{π x}{4})

x y d x + (1 + x^{2}) d y = 0

What is the integral of 1/(x^2*sqrt(5+x^2)) ?
The integral of 1/(x^2*sqrt(5+x^2)) is -(sqrt(5+x^2))/(5x)+C