What is the integral of sqrt(3+x^2) ?

The integral of sqrt(3+x^2) is 3(1/6 xsqrt(3+x^2)+1/2 ln| 1/(sqrt(3))x+sqrt(1/3 (3+x^2))|)+C

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integral of sqrt(3+x^2)

\int 3 + x^{2} d x

3 (\frac{1}{6} x 3 + x^{2} + \frac{1}{2} ln \frac{1}{3} x + \frac{1}{3} (3 + x^{2})) + C

Solution steps

\int 3 + x^{2} d x

Apply Trigonometric Substitution

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

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

Take the constant out:

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

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

Apply Integral Reduction

= 3 (\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) ∣

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

Substitute back

u = arctan (\frac{1}{3} x)

= 3 \frac{sec ^{2} ( arctan ( \frac{1}{3} x ) ) sin ( arctan ( \frac{1}{3} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{3} x)) + sec (arctan (\frac{1}{3} x))

Simplify

3 \frac{sec ^{2} ( arctan ( \frac{1}{3} x ) ) sin ( arctan ( \frac{1}{3} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{3} x)) + sec (arctan (\frac{1}{3} x)) : 3 (\frac{1}{6} x 3 + x^{2} + \frac{1}{2} ln \frac{1}{3} x + \frac{1}{3} (3 + x^{2}))

= 3 (\frac{1}{6} x 3 + x^{2} + \frac{1}{2} ln \frac{1}{3} x + \frac{1}{3} (3 + x^{2}))

Add a constant to the solution

= 3 (\frac{1}{6} x 3 + x^{2} + \frac{1}{2} ln \frac{1}{3} x + \frac{1}{3} (3 + x^{2})) + C

\int (3 x^{2} - 9 x + 2) d x

\frac{d}{d x} (3 x + 9)

\int \frac{1 + cos ( x )}{1 - cos ( x )} d x

\frac{d}{d x} (arcsin (5 x))

\frac{3}{5} x^{\frac{5}{3}} - \frac{3}{4} x^{\frac{1}{3}} + 9

What is the integral of sqrt(3+x^2) ?
The integral of sqrt(3+x^2) is 3(1/6 xsqrt(3+x^2)+1/2 ln| 1/(sqrt(3))x+sqrt(1/3 (3+x^2))|)+C