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Evaluate $\int \frac{\sqrt{x+4}}{x} d x$

let $u=\sqrt{x+4} \Rightarrow u^{2}=x+4 \Rightarrow x=u^{2}-4 \Rightarrow d x=2 u d u$

* $\int \frac{\sqrt{x+4}}{x} d x=\int \frac{u}{u^{2}-4} 2 u d u$

$=2 \int \frac{u^{2}}{u^{2}-4} d u$

long division

$* \int \frac{\sqrt{x+4}}{x}=2 \int\left(1+\frac{4}{u^{2}-4}\right) d u$

$\int f(x)=2 \int 1 d u+2 \int \frac{4}{u^{2}-4} d u$

$f(x)=2 \cdot(u)+8 \int \frac{1}{u^{2}-4} d u$

$\int \frac{1}{u^{2}-4} d u=\int \frac{1}{(u-2)(u+2)} d u$

$\Rightarrow f(x)=\frac{1}{(u-2)(u+2)} = \frac{A}{u-2}+\frac{B}{u+2}$

$A=(u-2) \cdot\left.\frac{1}{(u-2) (u+2)}\right|_{u=2}=\frac{1}{4}$

$B=(u+2) \cdot\left.\frac{1}{(u-2)(u+2)}\right|_{u=-2}=\frac{1}{-4}$

$\int \frac{\sqrt{x+4}}{x} d x=2 \cdot(u)+8 \int \frac{d u}{u^{2}-4}$

$\int \frac{\sqrt{x+4}}{x} d x=2 \cdot(u)+8 \int \frac{d u}{u^{2}-4}$

$\int \frac{1}{u^{2}-4}du=\int \frac{\frac{1}{4}}{(u-2)}du+\int \frac{-\frac{1}{4}}{a+2} d u$

$\int \frac{1}{u^{2}-4}=\frac{1}{4} \cdot \ln |u-2|-\frac{1}{4} \ln |u+2|+c$

$\int \frac{\sqrt{x+4}}{x} d x=2 \cdot \sqrt{x+4}+8\left[\frac{1}{4} \ln |u-2|-\frac{1}{4}\ln| u+2 |+c\right]$

$\int \frac{\sqrt{x+u}}{x} d x=2 \sqrt{x+4}+2 \ln |\sqrt{x+4}-2|-\frac{1}{4} \ln |\sqrt{x+4}+2|+c$