What is the integral of sqrt((243x)/4+82) ?

The integral of sqrt((243x)/4+82) is 1/729 (243x+328)^{3/2}+C

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integral of sqrt((243x)/4+82)

\int \frac{243 x}{4} + 82 d x

\frac{1}{729} (243 x + 328)^{\frac{3}{2}} + C

Solution steps

\int \frac{243 x}{4} + 82 d x

Simplify

\frac{243 x}{4} + 82 : \frac{1}{2} 243 x + 328

= \int \frac{1}{2} 243 x + 328 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 243 x + 328 d x

Apply u-substitution

: \int \frac{u}{243} d u

= \frac{1}{2} \cdot \int \frac{u}{243} d u

Take the constant out:

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

= \frac{1}{2} \cdot \frac{1}{243} \cdot \int u d u

Apply the Power Rule

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

= \frac{1}{2} \cdot \frac{1}{243} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 243 x + 328

= \frac{1}{2} \cdot \frac{1}{243} \cdot \frac{2}{3} (243 x + 328)^{\frac{3}{2}}

Simplify

\frac{1}{2} \cdot \frac{1}{243} \cdot \frac{2}{3} (243 x + 328)^{\frac{3}{2}} : \frac{1}{729} (243 x + 328)^{\frac{3}{2}}

= \frac{1}{729} (243 x + 328)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{729} (243 x + 328)^{\frac{3}{2}} + C

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What is the integral of sqrt((243x)/4+82) ?
The integral of sqrt((243x)/4+82) is 1/729 (243x+328)^{3/2}+C