J-0-3-2x-2-x-1-x-1-dx-

Question Number 121117 by benjo_mathlover last updated on 05/Nov/20

J = \underset{0}{\int^{3}} \frac{{2 x}^{2} + x - 1}{\sqrt{x + 1}} dx ?

Answered by liberty last updated on 05/Nov/20

\Rightarrow {2 x}^{2} + x - 1 = (2 x - 1) (x + 1)

let \sqrt{x + 1} = ρ \Rightarrow x = ρ^{2} - 1

J = \underset{1}{\int^{2}} \frac{(2 ρ^{2} - 3) ρ^{2} (2 ρ d ρ)}{ρ}

J = \int_{1}^{2} (4 ρ^{4} - 6 ρ^{2}) d ρ = \frac{4}{5} (32 - 1) - 2 (8 - 1)

= \frac{124}{5} - 14 = \frac{124 - 70}{5} = \frac{54}{5}

Answered by Bird last updated on 05/Nov/20

J = \int_{0}^{3} \frac{2 x^{2} + x - 1}{\sqrt{x + 1}} d x w e d o t h e c h a n g e m e n t

\sqrt{x + 1} = t \Rightarrow x + 1 = t^{2} \Rightarrow x = t^{2} - 1 \Rightarrow

J = \int_{1}^{2} \frac{2 {(t^{2} - 1)}^{2} + t^{2} - 1 - 1}{t} (2 t) d t

= 2 \int_{1}^{2} (2 (t^{4} - 2 t^{2} + 1) + t^{2} - 2) d t

= 2 \int_{1}^{2} (2 t^{4} - 3 t^{2}) d t

= 4 \int_{1}^{2} t^{4} d t + 6 \int_{1}^{2} t^{2} d t

= \frac{4}{5} {[t^{5}]}_{1}^{2} + 2 {[t^{3}]}_{1}^{2}

= \frac{4}{5} {2^{5} - 1} + 2 {2^{3} - 1}

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