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

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

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integral of 3/2 (x+1)^{1/2}

\int \frac{3}{2} (x + 1)^{\frac{1}{2}} d x

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

Solution steps

\int \frac{3}{2} (x + 1)^{\frac{1}{2}} d x

Take the constant out:

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

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

Multiply by the conjugate of

x + 1 : \frac{x ^{2} - 1}{x - 1}

= \frac{3}{2} \cdot \int (\frac{x ^{2} - 1}{x - 1})^{\frac{1}{2}} d x

Apply u-substitution

: \int (u + 2)^{\frac{1}{2}} d u

= \frac{3}{2} \cdot \int (u + 2)^{\frac{1}{2}} d u

Apply u-substitution

: \int v^{\frac{1}{2}} d v

= \frac{3}{2} \cdot \int v^{\frac{1}{2}} d v

Apply the Power Rule

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

= \frac{3}{2} \cdot \frac{2}{3} v^{\frac{3}{2}}

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

x \to - \frac{1}{2} lim (\frac{( 6 x - 1 )}{x})

1 7^{x}

\frac{d}{d x} (π - x)

\frac{x}{x + 2}

y^{^{'}} = y^{2} + x y^{2}

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