How do you find the integral of $\frac{5}{\sqrt{9 x^{2} - 16}} d x$ $5/[sqrt(9x^2-16)] dx$ ?

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How do you find the integral of $\frac{5}{\sqrt{9 x^{2} - 16}} d x$ $5/[sqrt(9x^2-16)] dx$ ?

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Answer 1 · 2018-04-01T17:10:27

The answer is $= \frac{5}{3} ln ⎛ ⎝ ∣ ∣ ∣ ∣ \frac{3}{4} x + \sqrt{{(\frac{3}{4} x)}^{2} - 1} ∣ ∣ ∣ ∣ ⎞ ⎠ + C$ $=5/3ln(|3/4x+sqrt((3/4x)^2-1)|)+C$

Explanation:

Perform some simplification

$\frac{5}{\sqrt{9 x^{2} - 16}} = \frac{5}{4 \sqrt{({(\frac{3}{4} x)}^{2}) - 1}}$ $5/sqrt(9x^2-16)=5/(4sqrt (((3/4x)^2)-1))$

Let $\frac{3}{4} x = sec u$ $3/4x=secu$ , $\Rightarrow$ $=>$ , $\frac{3}{4} d x = sec u tan u d u$ $3/4dx=secutanudu$

$\sqrt{{(\frac{3}{4} x)}^{2} - 1} = \sqrt{{sec}^{2} u - 1} = tan u$ $sqrt((3/4x)^2-1)=sqrt(sec^2u-1)=tanu$

Therefore, the integral is

$I = \int \frac{5 d x}{\sqrt{9 x^{2} - 16}} = \frac{5}{4} \int \frac{\frac{4}{3} sec u tan u d u}{tan u}$ $I=int(5dx)/(sqrt(9x^2-16))=5/4int(4/3secutanudu)/(tanu)$

$= \frac{5}{3} \int sec u d u$ $=5/3intsecudu$

$= \frac{5}{3} \int \frac{sec u (sec u + tan u) d u}{sec u + tan u}$ $=5/3int(secu(secu+tanu)du)/(secu+tanu)$

Let $v = sec u + tan u$ $v=secu+tanu$ , $\Rightarrow$ $=>$ , $d v = ({sec}^{2} u + sec u tan u) d u$ $dv=(sec^2u+secutanu)du$

Therefore,

$I = \frac{5}{3} \int \frac{d v}{v}$ $I=5/3int(dv)/(v)$

$= \frac{5}{3} ln (v)$ $=5/3ln(v)$

$= \frac{5}{3} ln (sec u + tan u)$ $=5/3ln(secu+tanu)$

$= \frac{5}{3} ln ⎛ ⎝ ∣ ∣ ∣ ∣ \frac{3}{4} x + \sqrt{{(\frac{3}{4} x)}^{2} - 1} ∣ ∣ ∣ ∣ ⎞ ⎠ + C$ $=5/3ln(|3/4x+sqrt((3/4x)^2-1)|)+C$

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How do you find the integral of $\frac{5}{\sqrt{9 x^{2} - 16}} d x$ $5/[sqrt(9x^2-16)] dx$ ?

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Explanation:

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How do you find the integral of 5/[sqrt(9x^2-16)] dx5√9x2−16dx5/[sqrt(9x^2-16)] dx?

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How do you find the integral of $\frac{5}{\sqrt{9 x^{2} - 16}} d x$ $5/[sqrt(9x^2-16)] dx$ ?