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

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

Solution steps

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

Apply Trigonometric Substitution

: \int \frac{1}{49 sec ( u )} d u

= \int \frac{1}{49 sec ( 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{1}{sec ( u )} d u

Rewrite using trig identities

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

Use the common integral:

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

= \frac{1}{49} sin (u)

Substitute back

u = arctan (\frac{1}{7} x)

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

Simplify

\frac{1}{49} sin (arctan (\frac{1}{7} x)) : \frac{x}{49 ( 49 + x ^{2} ) ^{\frac{1}{2}}}

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

Add a constant to the solution

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

\frac{d}{d t} (A t e^{- t})

y^{^{'}} + 7 (\frac{y}{x}) = x

\frac{d}{d x} (\frac{10}{1 + e ^{- 3 x}})

\frac{d}{d x} (e^{2 x} \cdot 4)

y = x, y = \frac{1}{2} x, x = 25

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