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

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

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

integral

\frac{x}{( a + x ) ^{\frac{1}{2}}}

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

Solution steps

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

Apply u-substitution

: \int \frac{u - a}{u ^{\frac{1}{2}}} d u

= \int \frac{u - a}{u ^{\frac{1}{2}}} d u

Expand

\frac{u - a}{u ^{\frac{1}{2}}} : u^{\frac{1}{2}} - \frac{a}{u ^{\frac{1}{2}}}

= \int u^{\frac{1}{2}} - \frac{a}{u ^{\frac{1}{2}}} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int u^{\frac{1}{2}} d u - \int \frac{a}{u ^{\frac{1}{2}}} d u

\int u^{\frac{1}{2}} d u = \frac{2}{3} u^{\frac{3}{2}}

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

= \frac{2}{3} u^{\frac{3}{2}} - a \cdot 2 u^{\frac{1}{2}}

Substitute back

u = a + x

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

Add a constant to the solution

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

1 - cos (x) 1 + cos (x)

\frac{cos ^{3} ( x )}{sin ( x )}

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

2 + 3 cos (0.5 x)

v = \frac{1}{3 t + 2}

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