What is the integral of sqrt(9x^2+1) ?

The integral of sqrt(9x^2+1) is 1/2 xsqrt(1+9x^2)+1/6 ln|3x+sqrt(1+9x^2)|+C

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integral of sqrt(9x^2+1)

\int 9 x^{2} + 1 d x

\frac{1}{2} x 1 + 9 x^{2} + \frac{1}{6} ln 3 x + 1 + 9 x^{2} + C

Solution steps

\int 9 x^{2} + 1 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 + 9 x^{2} + \frac{1}{6} ln 3 x + 1 + 9 x^{2}

= \frac{1}{2} x 1 + 9 x^{2} + \frac{1}{6} ln 3 x + 1 + 9 x^{2}

Add a constant to the solution

= \frac{1}{2} x 1 + 9 x^{2} + \frac{1}{6} ln 3 x + 1 + 9 x^{2} + C

\int \frac{1}{x ^{2} - 4 x - 12} d x

\frac{20}{s ( s ^{2} + 4 )}

\int \frac{1}{1 + cos ( 3 x )} d x

\frac{\partial}{\partial x} (22 x y e^{- x^{2} - y^{2}})

- 0.1 x^{2} + 1.2 x + 98.4

What is the integral of sqrt(9x^2+1) ?
The integral of sqrt(9x^2+1) is 1/2 xsqrt(1+9x^2)+1/6 ln|3x+sqrt(1+9x^2)|+C