What is the integral of 5/(x^2sqrt(25+x^2)) ?

The integral of 5/(x^2sqrt(25+x^2)) is -(sqrt(25+x^2))/(5x)+C

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integral of 5/(x^2sqrt(25+x^2))

\int \frac{5}{x ^{2} 25 + x ^{2}} d x

- \frac{25 + x ^{2}}{5 x} + C

Solution steps

\int \frac{5}{x ^{2} 25 + x ^{2}} d x

Take the constant out:

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

= 5 \cdot \int \frac{1}{x ^{2} 25 + x ^{2}} d x

Apply Trigonometric Substitution

= 5 \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

= 5 \cdot \frac{1}{25} \cdot \int \frac{sec ( u )}{tan ^{2} ( u )} d u

Rewrite using trig identities

= 5 \cdot \frac{1}{25} \cdot \int \frac{cos ( u )}{sin ^{2} ( u )} d u

Apply u-substitution

= 5 \cdot \frac{1}{25} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

= 5 \cdot \frac{1}{25} (- \frac{1}{v})

Substitute back

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

Simplify

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

= - \frac{25 + x ^{2}}{5 x}

Add a constant to the solution

= - \frac{25 + x ^{2}}{5 x} + C

What is the integral of 5/(x^2sqrt(25+x^2)) ?
The integral of 5/(x^2sqrt(25+x^2)) is -(sqrt(25+x^2))/(5x)+C