What is the integral of 1/(b^2sqrt(25+b^2)) ?

The integral of 1/(b^2sqrt(25+b^2)) is -(sqrt(25+b^2))/(25b)+C

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integral of 1/(b^2sqrt(25+b^2))

\int \frac{1}{b ^{2} 25 + b ^{2}} d b

- \frac{25 + b ^{2}}{25 b} + C

Solution steps

\int \frac{1}{b ^{2} 25 + b ^{2}} d b

Apply Trigonometric Substitution

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

= \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

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

Rewrite using trig identities

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

Apply u-substitution

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

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

Apply the Power Rule

: - \frac{1}{v}

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

Substitute back

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

Simplify

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

= - \frac{25 + b ^{2}}{25 b}

Add a constant to the solution

= - \frac{25 + b ^{2}}{25 b} + C

\frac{d}{d x} (2 - 4 cos (x))

y^{^{'}} = \frac{x y}{1 - x ^{2}}

f (x) = cos (x^{3})

\frac{2 x}{x ^{2} + y ^{2}}

\frac{90 t ^{2}}{t ^{2} + 150}

What is the integral of 1/(b^2sqrt(25+b^2)) ?
The integral of 1/(b^2sqrt(25+b^2)) is -(sqrt(25+b^2))/(25b)+C