What is the integral of 1/((169+x^2)^{3/2)} ?

The integral of 1/((169+x^2)^{3/2)} is x/(169(169+x^2)^{1/2)}+C

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integral of 1/((169+x^2)^{3/2)}

\int \frac{1}{( 169 + x ^{2} ) ^{\frac{3}{2}}} d x

\frac{x}{169 ( 169 + x ^{2} ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{1}{( 169 + x ^{2} ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

: \int \frac{1}{169 sec ( u )} d u

= \int \frac{1}{169 sec ( u )} d u

Take the constant out:

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

= \frac{1}{169} \cdot \int \frac{1}{sec ( u )} d u

Rewrite using trig identities

= \frac{1}{169} \cdot \int cos (u) d u

Use the common integral:

\int cos (u) d u = sin (u)

= \frac{1}{169} sin (u)

Substitute back

u = arctan (\frac{1}{13} x)

= \frac{1}{169} sin (arctan (\frac{1}{13} x))

Simplify

\frac{1}{169} sin (arctan (\frac{1}{13} x)) : \frac{x}{169 ( 169 + x ^{2} ) ^{\frac{1}{2}}}

= \frac{x}{169 ( 169 + x ^{2} ) ^{\frac{1}{2}}}

Add a constant to the solution

= \frac{x}{169 ( 169 + x ^{2} ) ^{\frac{1}{2}}} + C

y = x^{6} + 3

\frac{d}{d x} (sin (x^{2}) + sin^{2} (x))

\frac{\partial ^{2}}{\partial z ^{2}} (x^{2} y^{4} e^{z} + 3 y^{3} e^{x z} + 2 e^{x yz} + 3 y)

y^{^{'}} = \frac{( y ^{4} + 3 )}{4 x y ^{3}}

\frac{d y}{d x} = x^{2} \cdot y^{2}

What is the integral of 1/((169+x^2)^{3/2)} ?
The integral of 1/((169+x^2)^{3/2)} is x/(169(169+x^2)^{1/2)}+C