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

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

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integral of sqrt(25+x^2)

\int 25 + x^{2} d x

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

Solution steps

\int 25 + x^{2} 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

x \to + 0 lim (\frac{5 ^{x} - 7 ^{x}}{x})

y^{^{'}} = - a \cdot y + b \cdot y

f (x) = x^{2} + 2 (3.11)

- 4 cos (6 t)

\int (5^{x}) d x

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