How do you evaluate $\int \frac{x}{\sqrt{1 + x^{2}}}$ $int x/sqrt(1+x^2)$ from $[- \sqrt{2}, 0]$ $[-sqrt2,0]$ ?

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Answer 1 · 2018-03-26T16:34:57

$1 - \sqrt{3}$ $1-sqrt3$

$\int_{- \sqrt{2}}^{0} \frac{x}{\sqrt{1 + x^{2}}} d x$ $int_-sqrt2^0x/sqrt(1+x^2)dx$

$\int_{- \sqrt{2}}^{0} x {(1 + x^{2})}^{- \frac{1}{2}} d x$ $int_-sqrt2^0x(1+x^2)^(-1/2)dx$

by inspection

$= {[{(1 + x^{2})}^{\frac{1}{2}}]}_{- \sqrt{2}}^{\frac{1}{2}}$ $=[(1+x^2)^(1/2)]_-sqrt2^0$

$= {[{(1 + x^{2})}^{\frac{1}{2}}]}^{0} - {[{(1 + x^{2})}^{\frac{1}{2}}]}_{- \sqrt{2}}^{\frac{1}{2}}$ $=[(1+x^2)^(1/2)]^0-[(1+x^2)^(1/2)]_-sqrt2$

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

$= 1 - \sqrt{3}$ $=1-sqrt3$

How do you evaluate int x/sqrt(1+x^2)∫x√1+x2int x/sqrt(1+x^2) from [-sqrt2,0][−√2,0][-sqrt2,0]?