What is the integral of sqrt(x^2+21) ?

The integral of sqrt(x^2+21) is 21(1/42 xsqrt(21+x^2)+1/2 ln| 1/(sqrt(21))x+sqrt(1/21 (21+x^2))|)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of sqrt(x^2+21)

\int x^{2} + 21 d x

21 (\frac{1}{42} x 21 + x^{2} + \frac{1}{2} ln \frac{1}{21} x + \frac{1}{21} (21 + x^{2})) + C

Solution steps

\int x^{2} + 21 d x

Apply Trigonometric Substitution

: \int 21 sec^{3} (u) d u

= \int 21 sec^{3} (u) d u

Take the constant out:

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

= 21 \cdot \int sec^{3} (u) d u

Apply Integral Reduction

= 21 (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} \cdot \int sec (u) d u)

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

= 21 (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

u = arctan (\frac{1}{21} x)

= 21 \frac{sec ^{2} ( arctan ( \frac{1}{21} x ) ) sin ( arctan ( \frac{1}{21} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{21} x)) + sec (arctan (\frac{1}{21} x))

Simplify

21 \frac{sec ^{2} ( arctan ( \frac{1}{21} x ) ) sin ( arctan ( \frac{1}{21} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{21} x)) + sec (arctan (\frac{1}{21} x)) : 21 (\frac{1}{42} x 21 + x^{2} + \frac{1}{2} ln \frac{1}{21} x + \frac{1}{21} (21 + x^{2}))

= 21 (\frac{1}{42} x 21 + x^{2} + \frac{1}{2} ln \frac{1}{21} x + \frac{1}{21} (21 + x^{2}))

Add a constant to the solution

= 21 (\frac{1}{42} x 21 + x^{2} + \frac{1}{2} ln \frac{1}{21} x + \frac{1}{21} (21 + x^{2})) + C

\int (x + 2)^{15} d x

\int \frac{1}{12 + x ^{2}} d x

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

\int \frac{x + 2}{e ^{x}} d x

\int \frac{( x )}{x ^{2} + 4} d x

What is the integral of sqrt(x^2+21) ?
The integral of sqrt(x^2+21) is 21(1/42 xsqrt(21+x^2)+1/2 ln| 1/(sqrt(21))x+sqrt(1/21 (21+x^2))|)+C