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

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

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

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

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

Solution steps

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

\int sin (x) + 9 d x

y = x, y = 5 x, y = - x + 2

\int_{0}^{2} x^{3} - x^{2} d x

\int 5 arctan (2 y) d y

\frac{\partial}{\partial x} (ln (2 + x^{2} + 2 y^{2}))

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