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

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

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

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

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

Solution steps

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

y^{^{'}} + 4 x^{3} y^{2} = 0

\frac{\partial}{\partial x} (x^{2} - 5 x^{2} y - 6 x y^{2})

\int \frac{x ^{2} - x}{1 + 3 x ^{2} - 2 x ^{3}} d x

y = (x^{2} + 1)

\int \frac{36 + ( ln ( 8 x ) ) ^{2}}{x} d x

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