What is the integral of (4x+1)^{1/2} ?

The integral of (4x+1)^{1/2} is 1/6 (4x+1)^{3/2}+C

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integral of (4x+1)^{1/2}

\int (4 x + 1)^{\frac{1}{2}} d x

\frac{1}{6} (4 x + 1)^{\frac{3}{2}} + C

Solution steps

\int (4 x + 1)^{\frac{1}{2}} d x

Apply u-substitution

: \int \frac{u ^{\frac{1}{2}}}{4} d u

= \int \frac{u ^{\frac{1}{2}}}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int u^{\frac{1}{2}} d u

Apply the Power Rule

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

= \frac{1}{4} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 4 x + 1

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

Simplify

\frac{1}{4} \cdot \frac{2}{3} (4 x + 1)^{\frac{3}{2}} : \frac{1}{6} (4 x + 1)^{\frac{3}{2}}

= \frac{1}{6} (4 x + 1)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{6} (4 x + 1)^{\frac{3}{2}} + C

\int_{\frac{1}{2}}^{2} \frac{2}{x} - \frac{1}{2} x d x

\int \frac{4 - 32 x}{81 - 16 x ^{2}} d x

x \to 3 lim (\frac{9 + x}{3 - x})

x \to \infty lim (8 x + 5)

f (x) = - x^{2} - 3 x + 4

What is the integral of (4x+1)^{1/2} ?
The integral of (4x+1)^{1/2} is 1/6 (4x+1)^{3/2}+C