Question

Integral indefinida √x^2+16 dx

Answer 1

Integral por substituição trigonométrica

Vamos usar o segundo caso

$\int\sqrt{{x}^{2}+16}dx$

faça

$x=4\tan(\theta) \\dx=4{\sec}^{2}(\theta) d\theta\\ \sqrt{{x}^{2}+16}=4\sec(\theta)$

$\int\sqrt{{x}^{2}+16}dx\\=\int (4\sec(\theta) 4{\sec}^{2}(\theta) d\theta$

$\int (4\sec(\theta) 4{\sec}^{2}(\theta) d\theta=16\int {\sec}^{3}(\theta)d\theta$

$\int {\sec}^{3}(\theta)d\theta\\=\frac{1}{2}(\sec(\theta).\tan(\theta) +ln|\sec(\theta)+\tan(\theta)|)+k$

$\int(\sqrt{{x}^{2}+16})dx\\=16.\frac{1}{2}(\frac{x\sqrt{{x}^{2}+16}}{16}+ln|\frac{\sqrt{{x}^{2}+16}+x}{4}|)+k$

$\int(\sqrt{{x}^{2}+16})dx\\=8(\frac{x\sqrt{{x}^{2}+16}}{16}+ln|\frac{\sqrt{{x}^{2}+16}+x}{4}|)+k$