Question

Qual a integral dx/x^2*√(x^2-16) ?

Answer 1

$\int\ {\dfrac{\sqrt{x^2-16}}{x^2}} \, dx\\\\\\\dfrac{x}{4}=\sec\theta\\\\x=4\sec\theta\\\\dx=4\sec\theta\tan\theta\ d\theta\\\\\\\int\ {\dfrac{16\sec\theta\tan\theta\sqrt{\tan^2\theta}}{16\sec^2\theta}} \, d\theta\\\\\\\ \int {\dfrac{\tan^2\theta}{\sec\theta}} \, d\theta =\int {\dfrac{\sin^2\theta}{\cos\theta}} \, d\theta=\int {\dfrac{1-\cos^2\theta}{\cos\theta}} \, d\theta=\int {\dfrac{1}{\cos\theta}-\dfrac{\cos^2\theta}{\cos\theta}} \, d\theta=\\\\ \int {\sec\theta-\cos\theta} \, d\theta=\ln(\tan\theta+\sec\theta)-\sin\theta+C=\\\\\\\boxed{\ln\left(\dfrac{\sqrt{x^2-16}}{4}+\dfrac{x}{4} \right )-\dfrac{\sqrt{x^2-16}}{x}+C}$

Answer 2

Resposta:

Explicação passo-a-passo:

$\int \frac{1}{x^2\sqrt{x^2-16}}dx$

X = sdec u

$=\int \frac{1}{16\sec \left(u\right)}du\\=\frac{1}{16}\cdot \int \frac{1}{\sec \left(u\right)}du\\=\frac{1}{16}\cdot \int \cos \left(u\right)du\\=\frac{1}{16}\sin \left(u\right)$

fazendo u = arc (sec (x/4))

$=\frac{1}{16}\sin \left(arcsec(\frac{1}{4}x\right)\right)\\\\=\frac{\sqrt{x^2-16}}{16x} + c$