Question

Sa se determine integralele nedefinite
Subpunctele :b,e,f,h,i

Answer 1

Răspuns:

$b)~\int\frac{2x^{3}-x^{4}}{\sqrt{x} } dx=\\\\=\int\frac{2x^{3}-x^{4}}{x^\frac{1}{2}}dx\\\\=\int\frac{2x^{3}}{x^\frac{1}{2}}dx-\int\frac{x^{4}}{x^\frac{1}{2} } dx\\\\=\int{2x^\frac{5}{2} dx-\int{x^\frac{7}{2} }dx\\\\\\\\=2\int{x^\frac{5}{2}}dx-\int{x^\frac{7}{2}\\\\\\=\frac{2x^\frac{7}{2}}{\frac{7}{2} } -\frac{x^\frac{9}{2}}{\frac{9}{2}} +C\\\\=\frac{4x^\frac{7}{2}}{7} -\frac{2x^\frac{9}{2}}{9}+C$

$e)~ ln\sqrt[3]{4} =ln(2^{2})^\frac{1}{3}=ln2^\frac{2}{3}=\frac{2}{3}ln2\\=\int(2^{x}\cdot\frac{2}{3}ln2-ln3\cdot9^{x})dx\\\\=\frac{2}{3}ln2\int2^{x}dx-ln3\int9^{x}dx\\\\=\frac{2}{3}ln2\cdot\frac{2^{x}}{ln2}-ln3\cdot\frac{9^{x}}{ln9}+C\\\\=\frac{2^{x+1}}{3}-ln3\cdot\frac{9^{x}}{ln3^{2}}+C\\\\=\frac{2^{x+1}}{3}-ln3\cdot\frac{9^{x}}{2ln3}+C\\\\=\frac{2^{x+1}}{3}-\frac{9^{x}}{2}+C$

$f)~ \int\frac{1}{3+x^{2}}dx-\int\frac{1}{\sqrt{3+x^{2}} }dx=\frac{1}{\sqrt3}arctg\frac{x}{\sqrt3}-ln(x+\sqrt{3+x^{2}}~)$

$h) \int(\frac{\sqrt{x^2+4}-1}{x^2+4})dx\\\\=\int(\frac{\sqrt{x^2+4} }{x^2+4} -\frac{1}{x^2+4} )dx\\\\=\int\frac{\sqrt{x^2+4} }{x^2+4}dx-\int\frac{1}{x^2+4}dx\\\\=\int\sqrt\frac{x^2+4}{(x^2+4)^{2}} dx-\int\frac{1}{x^2+4}dx\\\\\\=\int\frac{1}{\sqrt{x^2+4}}dx-\int\frac{1}{x^2+4}dx\\\\\\\\=ln(x+\sqrt{x^2+4})+\frac{1}{2}arctg\frac{x}{2}+C$

$i) =\int\frac{\sqrt{x^2-4} }{x^2-4} dx+\int\frac{4}{x^2-4} dx\\\\=\int\sqrt{\frac{x^2-4}{(x^2-4)^{2}} } dx+4\int\frac{1}{x^2-4}dx\\\\=\int\frac{1}{\sqrt{x^2-4}}dx+4\int\frac{1}{x^2-4}dx\\\\\\=ln|x+\sqrt{x^2-4} |+\frac{1}{4}ln|\frac{x-2}{x+2}|+C$