What is the integral of sqrt(1+3x^2) ?

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

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integral of sqrt(1+3x^2)

\int 1 + 3 x^{2} d x

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

Solution steps

\int 1 + 3 x^{2} d x

Apply Trigonometric Substitution

: \int \frac{1}{3} sec^{3} (u) d u

= \int \frac{1}{3} sec^{3} (u) d u

Take the constant out:

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

= \frac{1}{3} \cdot \int sec^{3} (u) d u

Apply Integral Reduction

= \frac{1}{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) ∣

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

Substitute back

u = arctan (3 x)

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

Simplify

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

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

Add a constant to the solution

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

\int 1 + sin (y) d y

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