What is the integral of 1/((x^2-49)^{3/2)} ?

The integral of 1/((x^2-49)^{3/2)} is -x/(49(x^2-49)^{1/2)}+C

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integral of 1/((x^2-49)^{3/2)}

\int \frac{1}{( x ^{2} - 49 ) ^{\frac{3}{2}}} d x

- \frac{x}{49 ( x ^{2} - 49 ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{1}{( x ^{2} - 49 ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{49 tan ^{2} ( u )} d u

= \int \frac{sec ( u )}{49 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}{49} \cdot \int \frac{sec ( u )}{tan ^{2} ( u )} d u

Rewrite using trig identities

= \frac{1}{49} \cdot \int \frac{cos ( u )}{sin ^{2} ( u )} d u

Apply u-substitution

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

= \frac{1}{49} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= \frac{1}{49} (- \frac{1}{v})

Substitute back

= \frac{1}{49} (- \frac{1}{sin ( arcsec ( \frac{1}{7} x ) )})

Simplify

\frac{1}{49} (- \frac{1}{sin ( arcsec ( \frac{1}{7} x ) )}) : - \frac{x}{49 ( x ^{2} - 49 ) ^{\frac{1}{2}}}

= - \frac{x}{49 ( x ^{2} - 49 ) ^{\frac{1}{2}}}

Add a constant to the solution

= - \frac{x}{49 ( x ^{2} - 49 ) ^{\frac{1}{2}}} + C

- y^{^{'}} + \frac{2 y}{x} = x^{3} y^{2}

t \to 0 lim (\frac{6}{1 + t})

f (x) = arcsin (x)

\int 1 + (2 x)^{2} d x

y^{^{''}} + 2 y^{^{'}} + y = 12 e - t

What is the integral of 1/((x^2-49)^{3/2)} ?
The integral of 1/((x^2-49)^{3/2)} is -x/(49(x^2-49)^{1/2)}+C