What is the integral of sqrt(10+x^2) ?

The integral of sqrt(10+x^2) is 10(1/20 xsqrt(10+x^2)+1/2 ln| 1/(sqrt(10))x+sqrt(1/10 (10+x^2))|)+C

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integral of sqrt(10+x^2)

\int 10 + x^{2} d x

10 (\frac{1}{20} x 10 + x^{2} + \frac{1}{2} ln \frac{1}{10} x + \frac{1}{10} (10 + x^{2})) + C

Solution steps

\int 10 + x^{2} d x

Apply Trigonometric Substitution

= \int 10 sec^{3} (u) d u

Take the constant out:

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

= 10 \cdot \int sec^{3} (u) d u

Apply Integral Reduction

= 10 (\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) ∣

= 10 (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

u = arctan (\frac{1}{10} x)

= 10 \frac{sec ^{2} ( arctan ( \frac{1}{10} x ) ) sin ( arctan ( \frac{1}{10} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{10} x)) + sec (arctan (\frac{1}{10} x))

Simplify

10 \frac{sec ^{2} ( arctan ( \frac{1}{10} x ) ) sin ( arctan ( \frac{1}{10} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{10} x)) + sec (arctan (\frac{1}{10} x)) : 10 (\frac{1}{20} x 10 + x^{2} + \frac{1}{2} ln \frac{1}{10} x + \frac{1}{10} (10 + x^{2}))

= 10 (\frac{1}{20} x 10 + x^{2} + \frac{1}{2} ln \frac{1}{10} x + \frac{1}{10} (10 + x^{2}))

Add a constant to the solution

= 10 (\frac{1}{20} x 10 + x^{2} + \frac{1}{2} ln \frac{1}{10} x + \frac{1}{10} (10 + x^{2})) + C

What is the integral of sqrt(10+x^2) ?
The integral of sqrt(10+x^2) is 10(1/20 xsqrt(10+x^2)+1/2 ln| 1/(sqrt(10))x+sqrt(1/10 (10+x^2))|)+C