What is the integral of sqrt(1/(4x)+9) ?

The integral of sqrt(1/(4x)+9) is 1/12 (6sqrt(x)sqrt(36x+1)+ln|6sqrt(x)+sqrt(36x+1)|)+C

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integral of sqrt(1/(4x)+9)

\int \frac{1}{4 x} + 9 d x

\frac{1}{12} (6 x 36 x + 1 + ln 6 x + 36 x + 1) + C

Solution steps

\int \frac{1}{4 x} + 9 d x

\frac{1}{4 x} + 9 = 9 (\frac{\frac{1}{4 x}}{9} + 1)

= \int 9 (\frac{\frac{1}{4 x}}{9} + 1) d x

Simplify

9 (\frac{\frac{1}{4 x}}{9} + 1) : \frac{1 + 36 x}{2 x}

= \int \frac{1 + 36 x}{2 x} d x

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1 + 36 x}{x} d x

Apply Integration By Parts

: 2 x 1 + 36 x - \int \frac{36 x}{1 + 36 x} d x

= \frac{1}{2} (2 x 1 + 36 x - \int \frac{36 x}{1 + 36 x} d x)

\int \frac{36 x}{1 + 36 x} d x = \frac{1}{6} (- ln 6 x + 36 x + 1 + 6 x 36 x + 1)

= \frac{1}{2} (2 x 1 + 36 x - \frac{1}{6} (- ln 6 x + 36 x + 1 + 6 x 36 x + 1))

Simplify

\frac{1}{2} (2 x 1 + 36 x - \frac{1}{6} (- ln 6 x + 36 x + 1 + 6 x 36 x + 1)) : \frac{1}{12} (6 x 36 x + 1 + ln 6 x + 36 x + 1)

= \frac{1}{12} (6 x 36 x + 1 + ln 6 x + 36 x + 1)

Add a constant to the solution

= \frac{1}{12} (6 x 36 x + 1 + ln 6 x + 36 x + 1) + C

y^{^{'}} = \frac{( x ^{4} + x ^{2} y ^{2} + y ^{4} )}{x ^{3} y}

t - 2

y = lo g_{6} (x)

x \to 2 lim (x^{2})

\int \frac{1 + y}{y} d y

What is the integral of sqrt(1/(4x)+9) ?
The integral of sqrt(1/(4x)+9) is 1/12 (6sqrt(x)sqrt(36x+1)+ln|6sqrt(x)+sqrt(36x+1)|)+C