What is the integral of 1/((25+x^2)^{3/2)} ?

The integral of 1/((25+x^2)^{3/2)} is x/(25(25+x^2)^{1/2)}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/((25+x^2)^{3/2)}

\int \frac{1}{( 25 + x ^{2} ) ^{\frac{3}{2}}} d x

\frac{x}{25 ( 25 + x ^{2} ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{1}{( 25 + x ^{2} ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

: \int \frac{1}{25 sec ( u )} d u

= \int \frac{1}{25 sec ( 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{1}{sec ( u )} d u

Rewrite using trig identities

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

Use the common integral:

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

= \frac{1}{25} sin (u)

Substitute back

u = arctan (\frac{1}{5} x)

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

Simplify

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

= \frac{x}{25 ( 25 + x ^{2} ) ^{\frac{1}{2}}}

Add a constant to the solution

= \frac{x}{25 ( 25 + x ^{2} ) ^{\frac{1}{2}}} + C

\int \frac{u - 1}{u + 1} d u

y^{^{'}} = x y

x \to 5 lim (\frac{x π}{3})

\frac{1}{x ^{2}} d x = e^{- 2 t} d t

What is the integral of 1/((25+x^2)^{3/2)} ?
The integral of 1/((25+x^2)^{3/2)} is x/(25(25+x^2)^{1/2)}+C