x-4-1-x-5-4x-3-dx-

Question Number 118834 by bramlexs22 last updated on 20/Oct/20

\int \frac{x^{4} + 1}{x^{5} + 4 x^{3}} d x

Answered by bobhans last updated on 20/Oct/20

S o l v e \int \frac{x^{4} + 1}{x^{5} + 4 x^{3}} d x .

s o l u t i o n .

ψ = \int \frac{x^{4} + 1}{x^{3} (x^{2} + 4)} d x = \int \frac{x + x^{- 3}}{x^{2} + 4} d x

ψ = \int \frac{x}{x^{2} + 4} d x + \int \frac{x^{- 3}}{x^{2} + 4} d x

ψ = \frac{1}{2} \ln (x^{2} + 4) + \int \frac{d x}{x^{3} (x^{2} + 4)}

s e c o n d i n t e g r a l ψ_{2} = \int \frac{d x}{x^{3} (x^{2} + 4)}

l e t x = 2 \tan α

ψ_{2} = \int \frac{2 \sec^{2} α d α}{8 \tan^{3} α (4 \sec^{2} α)} = \frac{1}{16} \int \frac{\cos^{2} α d (\sin α)}{\sin^{3} α}

ψ_{2} = \int (\frac{1 - \sin^{2} α}{\sin^{3} α}) d (\sin α) = \int (u^{- 3} - u^{- 1}) d u

= - \frac{1}{2 u^{2}} - \ln (u) + c = - \frac{1}{2 \sin^{2} x} - \ln (\sin x) + c

T h u s ψ = \frac{1}{2} \ln (x^{2} + 4) - \ln (\sin x) - \frac{1}{2} cosec^{2} x + c

ψ = \ln (\frac{\sqrt{x^{2} + 4}}{\sin x}) - \frac{cosec^{2} x}{2} + c

Answered by MJS_new last updated on 20/Oct/20

\int \frac{x^{4} + 1}{x^{3} (x^{2} + 4)} d x =

= \frac{17}{16} \int \frac{x}{x^{2} + 4} d x - \frac{1}{16} \int \frac{d x}{x} + \frac{1}{4} \int \frac{d x}{x^{3}} =

= \frac{17}{32} \ln (x^{2} + 4) - \frac{1}{16} \ln ∣ x ∣ - \frac{1}{8 x^{2}} + C

Commented by bobhans last updated on 20/Oct/20

t y p o s i r . x^{4} (x^{2} + 4) = x^{6} + 4 x^{4}

Commented by MJS_new last updated on 20/Oct/20

yes, thank you

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