What is the integral of sqrt(9+4x^2) ?

The integral of sqrt(9+4x^2) is 1/4 (2xsqrt(4x^2+9)+9ln((|2x+sqrt(4x^2+9)|)/3))+C

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integral of sqrt(9+4x^2)

\int 9 + 4 x^{2} d x

\frac{1}{4} (2 x 4 x^{2} + 9 + 9 ln (\frac{2 x + 4 x ^{2} + 9}{3})) + C

Solution steps

\int 9 + 4 x^{2} d x

Apply Trigonometric Substitution

: \int \frac{9}{2} sec^{3} (u) d u

= \int \frac{9}{2} sec^{3} (u) d u

Take the constant out:

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

= \frac{9}{2} \cdot \int sec^{3} (u) d u

Apply Integral Reduction

= \frac{9}{2} (\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{9}{2} (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

u = arctan (\frac{2}{3} x)

= \frac{9}{2} (\frac{sec ^{2} ( arctan ( \frac{2}{3} x ) ) sin ( arctan ( \frac{2}{3} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{2}{3} x)) + sec (arctan (\frac{2}{3} x)))

Simplify

\frac{9}{2} (\frac{sec ^{2} ( arctan ( \frac{2}{3} x ) ) sin ( arctan ( \frac{2}{3} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{2}{3} x)) + sec (arctan (\frac{2}{3} x))) : \frac{1}{4} (2 x 4 x^{2} + 9 + 9 ln (\frac{2 x + 4 x ^{2} + 9}{3}))

= \frac{1}{4} (2 x 4 x^{2} + 9 + 9 ln (\frac{2 x + 4 x ^{2} + 9}{3}))

Add a constant to the solution

= \frac{1}{4} (2 x 4 x^{2} + 9 + 9 ln (\frac{2 x + 4 x ^{2} + 9}{3})) + C

\frac{( s + 2 )}{s ^{2} + s + 1} - \frac{1}{s}

5.2 x + 2.3

\int \frac{d y}{y} d x

(e^{c o s (x)})^{^{'}}

y = 3 tan^{2} (7 x)

What is the integral of sqrt(9+4x^2) ?
The integral of sqrt(9+4x^2) is 1/4 (2xsqrt(4x^2+9)+9ln((|2x+sqrt(4x^2+9)|)/3))+C