What is the integral of sqrt(3+9x^2) ?

The integral of sqrt(3+9x^2) is (sqrt(3))/2 xsqrt(1+3x^2)+1/2 ln|sqrt(3)x+sqrt(1+3x^2)|+C

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integral of sqrt(3+9x^2)

\int 3 + 9 x^{2} d x

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

Solution steps

\int 3 + 9 x^{2} d x

Apply Trigonometric Substitution

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

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

Apply Integral Reduction

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

Substitute back

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

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{4}{sin ^{2} ( 2 x )} d x

x \to 2 lim (\frac{x}{x ^{2} + 4})

\frac{\partial}{\partial x} (ln (x - y))

\int \frac{cos ( x )}{sin ^{2} ( x ) + sin ( x )} d x

d y = (y + 2 x - 3) d x

What is the integral of sqrt(3+9x^2) ?
The integral of sqrt(3+9x^2) is (sqrt(3))/2 xsqrt(1+3x^2)+1/2 ln|sqrt(3)x+sqrt(1+3x^2)|+C