How do you find the anti derivative of $\frac{\sqrt{x} - 8}{\sqrt{x} - 4}$ $(sqrtx -8)/(sqrtx-4)$ ?

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How do you find the anti derivative of $\frac{\sqrt{x} - 8}{\sqrt{x} - 4}$ $(sqrtx -8)/(sqrtx-4)$ ?

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Answer 1 · 2015-09-07T07:05:32

$\int \frac{\sqrt{x} - 8}{\sqrt{x} - 4} d x = x - 8 \sqrt{x} - 32 ln ∣ ∣ \sqrt{x} - 4 ∣ ∣$ $int (sqrtx -8)/(sqrtx -4) dx=x-8sqrtx -32ln|sqrtx -4|$ +C

Explanation:

The expression is equivalent to $1 - \frac{4}{\sqrt{x} - 4}$ $1 -4/(sqrtx -4)$

$\int \frac{\sqrt{x} - 8}{\sqrt{x} - 4} d x$ $int (sqrtx -8)/(sqrtx -4) dx$ = $\int (1 - \frac{4}{\sqrt{x} - 4}) d x = x - 4 \int \frac{1}{\sqrt{x} - 4} d x$ $int (1 -4/(sqrtx -4) )dx= x -4int 1/(sqrtx -4) dx$

Now to integrate $\frac{1}{\sqrt{x} - 4} d x$ $1/(sqrtx -4) dx$ , let x =t^2, so that dx= 2tdt and so $\int \frac{1}{\sqrt{x} - 4} d x = \int \frac{2 t d t}{t - 4}$ $int 1/(sqrtx-4) dx= int (2tdt)/(t-4)$

= $2 \int 1 + \frac{4}{t - 4} d t$ $2int 1+4/(t-4) dt$ = 2(t+4 ln|t-4))= $2 \sqrt{x} + 8 ln ∣ ∣ \sqrt{x} - 4 ∣ ∣$ $2sqrtx +8ln|sqrtx-4|$

$\int \frac{\sqrt{x} - 8}{\sqrt{x} - 4} d x = x - 8 \sqrt{x} - 32 ln ∣ ∣ \sqrt{x} - 4 ∣ ∣$ $int (sqrtx -8)/(sqrtx -4) dx=x-8sqrtx -32ln|sqrtx -4|$ +C

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How do you find the anti derivative of $\frac{\sqrt{x} - 8}{\sqrt{x} - 4}$ $(sqrtx -8)/(sqrtx-4)$ ?

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Explanation:

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How do you find the anti derivative of $\frac{\sqrt{x} - 8}{\sqrt{x} - 4}$ $(sqrtx -8)/(sqrtx-4)$ ?