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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{4 tan ^{2} ( u )} d u

= 4 \cdot \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

= 4 \cdot \frac{1}{4} \cdot \int \frac{sec ( u )}{tan ^{2} ( u )} d u

Rewrite using trig identities

= 4 \cdot \frac{1}{4} \cdot \int \frac{cos ( u )}{sin ^{2} ( u )} d u

Apply u-substitution

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

= 4 \cdot \frac{1}{4} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= 4 \cdot \frac{1}{4} (- \frac{1}{v})

Substitute back

= 4 \cdot \frac{1}{4} (- \frac{1}{sin ( arctan ( \frac{1}{2} x ) )})

Simplify

4 \cdot \frac{1}{4} (- \frac{1}{sin ( arctan ( \frac{1}{2} x ) )}) : - \frac{4 + x ^{2}}{x}

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

Add a constant to the solution

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

2 y \frac{d y}{d x} + 8 x = 6

ln (\frac{2}{3} x)

\frac{d}{d x} (20 x^{2} + \frac{180}{x})

(y - x) \frac{d y}{d x} = 1

\frac{d y}{d θ} = 4 y^{2} sec^{2} (2 θ)

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