What is the integral of (r^2)/(sqrt(1+r^2)) ?

The integral of (r^2)/(sqrt(1+r^2)) is 1/2 rsqrt(r^2+1)-1/2 ln|r+sqrt(r^2+1)|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (r^2)/(sqrt(1+r^2))

integral

\frac{r ^{2}}{1 + r ^{2}}

\frac{1}{2} r r^{2} + 1 - \frac{1}{2} ln r + r^{2} + 1 + C

Solution steps

\int \frac{r ^{2}}{1 + r ^{2}} d r

Apply Trigonometric Substitution

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

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

Rewrite using trig identities

= \int (- 1 + sec^{2} (u)) sec (u) d u

Expand

(- 1 + sec^{2} (u)) sec (u) : - sec (u) + sec^{3} (u)

= \int - sec (u) + sec^{3} (u) d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int sec (u) d u + \int sec^{3} (u) d u

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

\int sec^{3} (u) d u = \frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣

= - ln ∣ tan (u) + sec (u) ∣ + \frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣

Substitute back

u = arctan (r)

= - ln ∣ tan (arctan (r)) + sec (arctan (r)) ∣ + \frac{1}{2} sec (arctan (r)) tan (arctan (r)) + \frac{1}{2} ln ∣ tan (arctan (r)) + sec (arctan (r)) ∣

Simplify

- ln ∣ tan (arctan (r)) + sec (arctan (r)) ∣ + \frac{1}{2} sec (arctan (r)) tan (arctan (r)) + \frac{1}{2} ln ∣ tan (arctan (r)) + sec (arctan (r)) ∣ : \frac{1}{2} r r^{2} + 1 - \frac{1}{2} ln r + r^{2} + 1

= \frac{1}{2} r r^{2} + 1 - \frac{1}{2} ln r + r^{2} + 1

Add a constant to the solution

= \frac{1}{2} r r^{2} + 1 - \frac{1}{2} ln r + r^{2} + 1 + C

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

\frac{x}{x ^{2}}

\frac{e ^{x}}{( x - 1 )}

36 - 25 x^{2}

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

What is the integral of (r^2)/(sqrt(1+r^2)) ?
The integral of (r^2)/(sqrt(1+r^2)) is 1/2 rsqrt(r^2+1)-1/2 ln|r+sqrt(r^2+1)|+C