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Find the anti-derivative of xsqrtx+1 using substitution show steps
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are u sure its \(f(x)=x\sqrt x+1\) ?
\[x \sqrt{x+1}\]
ok, then put x+1 = u find du=.. ?
du=((x^2)/2)+x ?
x+1 = u taking deribative dx+0 = du so, dx is just du.
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oooo I was doing the anti derivative for du. ok thanks
hmm....so could you solove further ?
so now the equation is sqrtu du ?
\(\sqrt{x+1} = \sqrt u \\ x=u-1 \\dx=du\) so whats your new integral ?
u-1sqrtu du
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yes, now thats an easy integral t solve can u ?
yea I got this thanks
welcome ^_^
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