1-8-x-2-4-x-2-dx-2-x-6-2x-2-5x-3-

Question Number 136968 by Eric002 last updated on 28/Mar/21

1) \int \frac{8}{x^{2} \sqrt{4 - x^{2}}} d x

2) \int \frac{x - 6}{2 x^{2} - 5 x + 3}

Answered by EDWIN88 last updated on 28/Mar/21

(1) \int \frac{8}{x^{3} \sqrt{{4 x}^{- 2} - 1}} dx = \int \frac{{8 x}^{- 3}}{\sqrt{{4 x}^{- 2} - 1}} dx

set u^{2} = {4 x}^{- 2} - 1 2 u du = - {8 x}^{- 3} dx

E = - 2 \int \frac{u}{u} du = - 2 u + c

E = - 2 \sqrt{{4 x}^{- 2} - 1} + c = - 2 \sqrt{\frac{4 - x^{2}}{x^{2}}} + c

E = - \frac{2 \sqrt{4 - x^{2}}}{x} + c

Answered by EDWIN88 last updated on 28/Mar/21

(2) E = \int \frac{x - 6}{(2 x + 1) (x - 3)} dx

Partial fraction \frac{x - 6}{{2 x}^{2} - 5 x + 3} = \frac{p}{2 x + 1} + \frac{q}{x - 3}

where {\begin{cases} p = [\frac{x - 6}{x - 3}]_{x = - \frac{1}{2}} = \frac{- 13}{- 7} = \frac{13}{7} \\ q = {[\frac{x - 6}{2 x + 1}]}_{x = 3} = - \frac{3}{7} \end{cases}

E = \frac{13}{7} \int \frac{dx}{2 x + 1} - \frac{3}{7} \int \frac{dx}{x - 3}

E = \frac{13}{14} \ln ∣ 2 x + 1 ∣ - \frac{3}{7} \ln ∣ x - 3 ∣ + c

Answered by mathmax by abdo last updated on 28/Mar/21

1) I = \int \frac{8}{x^{2} \sqrt{4 - x^{2}}} dx \Rightarrow I =_{x = 2 \sin θ} \int \frac{8}{{4 \sin}^{2} θ . 2 \cos θ} (2 \cos θ) d θ

= 4 \int \frac{d θ}{\sin^{2} θ} = 8 \int \frac{d θ}{1 - \cos (2 θ)} =_{2 θ = t} 8 \int \frac{dt}{2 (1 - cost)} = 4 \int \frac{dt}{1 - cost}

=_{\tan (\frac{t}{2}) = y} 4 \int \frac{2 dy}{(1 + y^{2}) (1 - \frac{1 - y^{2}}{1 + y^{2}})} = 8 \int \frac{dy}{1 + y^{2} - 1 + y^{2}} = 4 \int \frac{dy}{y^{2}}

= - \frac{4}{y} + C = - \frac{4}{\tan (\frac{t}{2})} + C = - \frac{4}{\tan (θ)} + C = - \frac{4}{\tan (\arcsin (\frac{x}{2}))} + C