Question

Calcule o valor das integrais
dica: use a formula: ∫$\frac{1}{\sqrt{a^{2} +x^{2} } } dx=ln |x+\sqrt{a^{2} +x^{2} |} +k$

Answer 1

Resposta:

$\Large Dica : ∫\frac{1}{\sqrt{a^{2} +x^{2} } } dx=ln |x+\sqrt{a^{2} +x^{2} |} +k \\ \\ \Large \bf Utilizando \: a \: dica\: temos:$

a)

$\large \: \int \frac{1}{ \sqrt{ {x}^{2} + \frac{9}{25} } } dx = \\ \\ = \int \frac{1}{ \sqrt{ {x}^{2} + \left( \frac{3}{5} \right) {}^{2} }}dx \\ \\ = ln \left |x + \sqrt{ {x}^{2} + \frac{9}{25} } \right | + k \\ \\ = ln \left |x + \sqrt{ \frac{ 25{x}^{2} + 9 }{25} } \right | + k \\ \\ = ln \left |x + \frac{ \sqrt{25 {x}^{2} + 9 } }{ \sqrt{25} } \right | + k \\ \\ = \Large \boxed{ \green{ln \left|x + \dfrac{ \sqrt{ {x}^{2} + 25 } }{5} \right| + k}}$

b)

$\large \int \frac{1}{ \sqrt{ {x}^{2} + 49} } dx = \\ \\ = \int \frac{1}{ \sqrt{ {x}^{2} + {7}^{2} } } dx \\ \\ = \Large \boxed{ \green{ln \left|x + \sqrt{ {x}^{2} + 49} \right| + k}}\\ \\ \Large \boxed{\underline{\blue{\bf Bons\: Estudos!}\:\bf 17/05/2021}}$

Answer 2

∫ 1/√(1/16+x²)   dx

Fazendo a substituição

x=(1/4)*tg(u)  ==> dx=(1/4)* sec²(u) du

∫ { 1/√[1/16+(1/16) *tg²(u)] }  *     (1/4)* sec²(u)  du

∫ { 1/(1/4)*√[1+tg²(u)] }  *     (1/4)* sec²(u)  du

∫ { 1/√[1+(tg²(u)] }  *     sec²(u)  du

∫ { 1/√[cos²(u)/cos²(u)+sen²(u)/cos²(u)] }  *     sec²(u)  du

∫ { 1/√[1/cos²(u)] }  *     sec²(u)  du

∫ { 1/√[sec²(u)] }  *     sec²(u)  du

∫  1/sec (u)  *     sec²(u)  du

multiplique por (sec(u)+tan(u))/(sec(u)+tan(u))

∫   sec (u)*(sec(u)+tan(u))/(sec(u)+tan(u))   du

∫ (sec²(u)+sec(u) *tan(u))/(sec(u)+tan(u))   du

Substitua

s=tan(u) +sec(u) ==>ds=sec²(u)+tan(s)*sec(u) du

∫ (sec²(u)+sec(u) *tan(u))/s)   ds /(sec²(u)+tan(s)*sec(u) )

∫ 1/s   ds  = ln |s| + c

Como s=tan(u) +sec(u)

= ln |tan(u) +sec(u)| + c

Como x=(1/4)*tg(u)  ==> u =arctan(4x)

= ln |tan(arctan(4x)) +sec(arctan(4x))| + c  é a resposta