Solved: "How do you solve the integral of xx+8−xdx?"

Devin Dougherty

Answered question

2022-04-09

How do you solve the integral of

\frac{\sqrt{x}}{\sqrt{x} + \sqrt{8 - x}} d x

embemiaEffoset4rs

Beginner2022-04-10Added 14 answers

We are evaluating the integral which follows:

I = \int \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a - x}} \partial x

We begin with rationalisation,

I = \int \sqrt{x} \cdot \frac{\sqrt{x} - \sqrt{a - x}}{x - (a - x)} \partial x

which separate it into two:

I = \frac{1}{2} (I_{p} - I_{q})

I_{p} = \int \frac{x}{x - \frac{a}{2}} \partial x

I_{q} = \int \frac{\sqrt{x (a - x)}}{x - \frac{a}{2}} \partial x

We begin with the first one,

I_{p} = \int (1 + \frac{\frac{a}{2}}{x - \frac{a}{2}}) \partial x

= x + \frac{a}{2} \ln | x - \frac{a}{2} | + C_{p}

which is easier. The second half follows:

I_{q} = \int \frac{\sqrt{\frac{a^{2}}{4} - {(x - \frac{a}{2})}^{2}}}{x - \frac{a}{2}} \partial x

We’ve completed the square. Set

2 x - a = a \sin (θ)

I_{q} = \frac{a}{2} \int \frac{\cos^{2} (θ)}{\sin (θ)} \partial θ

= \frac{a}{2} \int (\frac{1}{\sin (θ)} - \sin (θ)) \partial θ

= \frac{a}{2} \cos (θ) + \frac{a}{2} \int \frac{\sin (θ)}{1 - \cos^{2} (θ)} \partial θ

= \frac{a}{2} \cos (θ) + \frac{a}{4} \int (\frac{\sin (θ)}{1 - \cos (θ)} + \frac{\sin (θ)}{1 + \cos (θ)}) \partial θ

= \frac{a}{2} \cos (θ) + \frac{a}{4} \ln | \frac{1 - \cos (θ)}{1 + \cos (θ)} | + C_{q}

Finally, put everything together:

I = \frac{1}{2} (x - \sqrt{x (a - x)}) + \frac{a}{4} \ln | x - \frac{a}{2} | + \frac{a}{8} \ln | \frac{\frac{a}{2} + \sqrt{x (a - x)}}{\frac{a}{2} - \sqrt{x (a - x)}} | + C

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