What is the integral of (-28)/(x^2sqrt(x^2+25)) ?

The integral of (-28)/(x^2sqrt(x^2+25)) is (28sqrt(25+x^2))/(25x)+C

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integral of (-28)/(x^2sqrt(x^2+25))

\int \frac{- 28}{x ^{2} x ^{2} + 25} d x

\frac{28 25 + x ^{2}}{25 x} + C

Solution steps

\int \frac{- 28}{x ^{2} x ^{2} + 25} d x

Take the constant out:

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

= - 28 \cdot \int \frac{1}{x ^{2} x ^{2} + 25} d x

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{25 tan ^{2} ( u )} d u

= - 28 \cdot \int \frac{sec ( u )}{25 tan ^{2} ( u )} d u

Take the constant out:

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

= - 28 \cdot \frac{1}{25} \cdot \int \frac{sec ( u )}{tan ^{2} ( u )} d u

Rewrite using trig identities

= - 28 \cdot \frac{1}{25} \cdot \int \frac{cos ( u )}{sin ^{2} ( u )} d u

Apply u-substitution

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

= - 28 \cdot \frac{1}{25} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= - 28 \cdot \frac{1}{25} (- \frac{1}{v})

Substitute back

= - 28 \cdot \frac{1}{25} (- \frac{1}{sin ( arctan ( \frac{1}{5} x ) )})

Simplify

- 28 \cdot \frac{1}{25} (- \frac{1}{sin ( arctan ( \frac{1}{5} x ) )}) : \frac{28 25 + x ^{2}}{25 x}

= \frac{28 25 + x ^{2}}{25 x}

Add a constant to the solution

= \frac{28 25 + x ^{2}}{25 x} + C

x \to 10 lim (\frac{1}{x - 10})

n = 1 \sum \infty (- \frac{3}{2})^{n}

\int_{- 1}^{0} x - 2 d x

\int \frac{1}{x ( x + 1 ) ( x - 2 )} d x

\frac{d y}{d x} = \frac{y}{x} + y^{4}

What is the integral of (-28)/(x^2sqrt(x^2+25)) ?
The integral of (-28)/(x^2sqrt(x^2+25)) is (28sqrt(25+x^2))/(25x)+C