∫ [tex] \frac{x+2}{\sqrt{ x^{2} +4x-3} } [/tex] dx tolong ya kawan kawan

jawab

u = x² + 4x  - 3

du = 2x + 4 dx
du = 2(x +2) dx
(x+2) dx = 1/2 du

[tex] \int{ \frac{x+2}{ \sqrt{x^{2}+4x - 3} } \, dx [/tex]

[tex]\int{ (x+2)(x^{2} +4x - 3)^{- \frac{1}{2}}\, dx[/tex]

[tex]\int{ \frac{1}{2}(u)^{- \frac{1}{2}}\, dx [/tex]

= U^(1/2) + c

= √(x² + 4x - 3) + c

[tex] \frac {1}{2}\int\limits (x^2+4x-3)^{\frac {-1}{2}} \ d(x^2+4x-3)\\ \frac {1}{2} \int\limits u^{\frac {-1}{2}}\ du\\ \frac {1}{2}.2\sqrt u+c\\ \sqrt u+c\\ \sqrt{ x^2+4x-3}\ + C \\ \\ CATATAN:\\ \\ d(x^2+4x-3)=2x+4\ dx\\ ,\ lalu\ dikalikan\ \frac {1}{2}\ agar\ sama\ dengan\ soal\ yaitu\ menjadi\ x+2[/tex]