What is the integral of sqrt(x^2+25) ?

The integral of sqrt(x^2+25) is 25(1/50 xsqrt(25+x^2)+1/2 ln((|x+sqrt(25+x^2)|)/5))+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+25)

\int x^{2} + 25 d x

25 (\frac{1}{50} x 25 + x^{2} + \frac{1}{2} ln (\frac{x + 25 + x ^{2}}{5})) + C

Solution steps

\int x^{2} + 25 d x

Apply Trigonometric Substitution

: \int 25 sec^{3} (u) d u

= \int 25 sec^{3} (u) d u

Take the constant out:

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

= 25 \cdot \int sec^{3} (u) d u

Apply Integral Reduction

= 25 (\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) ∣

= 25 (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

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

= 25 (\frac{sec ^{2} ( arctan ( \frac{1}{5} x ) ) sin ( arctan ( \frac{1}{5} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{5} x)) + sec (arctan (\frac{1}{5} x)))

Simplify

25 (\frac{sec ^{2} ( arctan ( \frac{1}{5} x ) ) sin ( arctan ( \frac{1}{5} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{5} x)) + sec (arctan (\frac{1}{5} x))) : 25 (\frac{1}{50} x 25 + x^{2} + \frac{1}{2} ln (\frac{x + 25 + x ^{2}}{5}))

= 25 (\frac{1}{50} x 25 + x^{2} + \frac{1}{2} ln (\frac{x + 25 + x ^{2}}{5}))

Add a constant to the solution

= 25 (\frac{1}{50} x 25 + x^{2} + \frac{1}{2} ln (\frac{x + 25 + x ^{2}}{5})) + C

\int \frac{3}{9 - x ^{2}} d x

\frac{d}{d x} ((3 x^{6} + 8 x^{3})^{4})

\frac{d}{d x} (\frac{2}{4 x + 3})

5 sec (x)

\frac{d}{d x} (\frac{2 x}{x ^{2} + y ^{2}})

What is the integral of sqrt(x^2+25) ?
The integral of sqrt(x^2+25) is 25(1/50 xsqrt(25+x^2)+1/2 ln((|x+sqrt(25+x^2)|)/5))+C