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Grade 11Algebra

Let f(x)=integral of (2u+1)÷(u²-2u+2) du then find the atleast value of (x) in the interval[-1,1] {upperlimit is x, lower limit is 0}

max1234567890

12 Years agoGrade 11

1 Answer

Sher Mohammad

12 Years ago

$(u-2)^{2}-u^{2}-1=4u$

$\frac{1}2{}\frac{4u+2}{u^2-2u+2}=\frac{1}2\frac{(u-2)^{2}-u^{2}-1+2}{(u-2)^{2}-2}=\frac{1}2\frac{(u-2)^{2}-u^{2}+1}{(u-2)^{2}-2}$

$\frac{1}2{}\frac{4u+2}{u^2-2u+2}=\frac{1}2\frac{(u-2)^{2}-2-u^{2}+3}{(u-2)^{2}-2}=\frac{1}{2}+\frac{3-u^{2}}{(u-2)^{2}-2}$

$\frac{1}2{}\frac{4u+2}{u^2-2u+2}=\frac{1}{2}+\frac{3}{(u-2)^{2}-2}-\frac{u^{2}}{(u-2)^{2}-2}$

$\int \frac{3}{(u-2)^{2}-2}=\frac{3 \left(\log \left(-u+\sqrt{2}+2\right)-\log \left(u+\sqrt{2}-2\right)\right)}{2 \sqrt{2}}$

$\int -\frac{u^{2}}{(u-2)^{2}-2}=-u-\left(2+\frac{3}{\sqrt{2}}\right) \log \left(-u+\sqrt{2}+2\right)-\left(2-\frac{3}{\sqrt{2}}\right) \log \left(u+\sqrt{2}-2\right)$

$\int \frac{1}{2}=\frac{1}{2}u$

Sher Mohammad,

B.Tech, IIT Delhi

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