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

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

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

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

Solution steps

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

Apply Trigonometric Substitution

: \int \frac{1}{a ^{2} sec ( u )} d u

= \int \frac{1}{a ^{2} sec ( u )} d u

Take the constant out:

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

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

Rewrite using trig identities

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

Use the common integral:

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

= \frac{1}{a ^{2}} sin (u)

Substitute back

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

= \frac{1}{a ^{2}} sin (arctan (\frac{1}{a} x))

Simplify

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

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

Add a constant to the solution

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

(x + 4) (x^{2} - 4)

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

x^{2} - 1, [- 1, 2]

\frac{d ^{2}}{d x ^{2}} (2 csc (x))

x \to 4 lim (x^{2} - 5)

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