What is $\int \frac{x}{\sqrt{x^{2} - 8^{2}}} d x$ $int x/ sqrt(x^2 - 8^2) dx$ ?

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Answer 1 · 2016-02-04T13:12:04

$\int \frac{x}{\sqrt{x^{2} - 64}} d x = \sqrt{x^{2} - 64} + C$ $int x/(sqrt(x^2 - 64)) "d"x = sqrt(x^2 - 64) + C$

where $C$ $C$ is a constant.

$\int \frac{x}{\sqrt{x^{2} - 8^{2}}} d x = \int \frac{x}{\sqrt{x^{2} - 64}} d x$ $int x/(sqrt(x^2 - 8^2)) "d"x = int x/(sqrt(x^2 - 64)) "d"x$

Let $u = \sqrt{x^{2} - 64}$ $u = sqrt(x^2 - 64)$ . We need to differentiate $u$ $u$ as next:

$("d"u)/("d"x) = 1 / (2 sqrt(x^2 - 64)) * 2x = (cancel(2)x)/(cancel(2) sqrt(x^2 - 64)) = x / (sqrt(x^2 - 64))$

Multiply by $"d"x$ to have just $"d"u$ on one side:

$"d"u = x / (sqrt(x^2 - 64)) "d"x$

As you can see, the substition seems to fit our integral perfectly. So we can start to substitute and solve the integral:

$int x/(sqrt(x^2 - 64)) "d"x = int 1 * color(blue)(x/(sqrt(x^2 - 64)) "d"x)$

... replace $x / (sqrt(x^2 - 64)) "d"x$ by $"d"u$ and all other occurences of $sqrt(x^2 - 64)$ , if present, by $u$ ...

$= int 1 color(white)(ii) color(blue)("d"u)$

... the integral of $1$ is $u + C$ with a constant $C$ ...

$= u + C$

... re-substitute by replacing $u$ with $sqrt(x^2 - 64)$ ...

$= sqrt(x^2 - 64) + C$

Hope that this helped!

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