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

The integral of sqrt(1+9/4 x) is 1/27 (4+9x)^{3/2}+C

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integral of sqrt(1+9/4 x)

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

\frac{1}{27} (4 + 9 x)^{\frac{3}{2}} + C

Solution steps

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

Apply u-substitution

: \int \frac{4 u}{9} d u

= \int \frac{4 u}{9} d u

Take the constant out:

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

= \frac{4}{9} \cdot \int u d u

Apply the Power Rule

: \frac{2}{3} u^{\frac{3}{2}}

= \frac{4}{9} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 1 + \frac{9}{4} x

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

Simplify

\frac{4}{9} \cdot \frac{2}{3} (1 + \frac{9}{4} x)^{\frac{3}{2}} : \frac{1}{27} (4 + 9 x)^{\frac{3}{2}}

= \frac{1}{27} (4 + 9 x)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{27} (4 + 9 x)^{\frac{3}{2}} + C

\int 1 + x^{3} d x

f (x) = \frac{x ^{2}}{x - 3}

e^{2 x} cos (3 x)

f (x) = 4 tan (x), at x = \frac{π}{3}

\int_{0}^{3} π (9 - x^{2})^{2} d x

What is the integral of sqrt(1+9/4 x) ?
The integral of sqrt(1+9/4 x) is 1/27 (4+9x)^{3/2}+C