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

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

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

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

5 π^{6}

x \to 0 - lim (arctan (ln (x)))

\frac{\partial}{\partial y} (tan (2 + 2 x^{2} y^{3} z^{2}))

\int x^{\frac{7}{6}} d x

\int_{- 1}^{2} 5 d x

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