What is the integral of (-15)/(x^2sqrt(x^2+4)) ?

The integral of (-15)/(x^2sqrt(x^2+4)) is (15sqrt(4+x^2))/(4x)+C

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integral of (-15)/(x^2sqrt(x^2+4))

\int \frac{- 15}{x ^{2} x ^{2} + 4} d x

\frac{15 4 + x ^{2}}{4 x} + C

Solution steps

\int \frac{- 15}{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

= - 15 \cdot \int \frac{1}{x ^{2} x ^{2} + 4} d x

Apply Trigonometric Substitution

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

= - 15 \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

= - 15 \cdot \frac{1}{4} \cdot \int \frac{sec ( u )}{tan ^{2} ( u )} d u

Rewrite using trig identities

= - 15 \cdot \frac{1}{4} \cdot \int \frac{cos ( u )}{sin ^{2} ( u )} d u

Apply u-substitution

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

= - 15 \cdot \frac{1}{4} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= - 15 \cdot \frac{1}{4} (- \frac{1}{v})

Substitute back

= - 15 \cdot \frac{1}{4} (- \frac{1}{sin ( arctan ( \frac{1}{2} x ) )})

Simplify

- 15 \cdot \frac{1}{4} (- \frac{1}{sin ( arctan ( \frac{1}{2} x ) )}) : \frac{15 4 + x ^{2}}{4 x}

= \frac{15 4 + x ^{2}}{4 x}

Add a constant to the solution

= \frac{15 4 + x ^{2}}{4 x} + C

\int 3 \frac{x ^{4}}{e ^{x^{5}}} d x

\int_{0}^{\frac{π}{4}} 1

9 x + 1 - x^{2}

x \to 1 lim (\frac{x ^{2} - 3 x}{x - 1})

y = (x^{2} + 4 x - 3)^{4}

What is the integral of (-15)/(x^2sqrt(x^2+4)) ?
The integral of (-15)/(x^2sqrt(x^2+4)) is (15sqrt(4+x^2))/(4x)+C