How do you use substitution to integrate $x \sqrt{2 x - 1}$ $xsqrt(2x-1)$ ?

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How do you use substitution to integrate $x \sqrt{2 x - 1}$ $xsqrt(2x-1)$ ?

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Answer 1 · 2015-10-10T14:50:15

Set $u = 2 x - 1 \Rightarrow x = \frac{u + 1}{2}$ $u=2x-1=>x=(u+1)/2$ hence $d u = 2 d x$ $du=2dx$ so now we have

$\int x \sqrt{2 x - 1} d x = \int \frac{u + 1}{2} \cdot \sqrt{u} \cdot \frac{d u}{2} = \frac{1}{4} \int (u \sqrt{u} + \sqrt{u}) d u = \frac{1}{4} \cdot \frac{2}{15} \cdot u^{\frac{3}{2}} \cdot (3 u + 5) + c = \frac{1}{15} \cdot {(2 x - 1)}^{\frac{3}{2}} \cdot (3 x + 1) + c$ $int xsqrt(2x-1)dx=int (u+1)/2*sqrtu*(du)/2=1/4 int (usqrtu+sqrtu)du= 1/4*2/15*u^(3/2)*(3u+5)+c=1/15*(2x-1)^(3/2)*(3x+1)+c$

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How do you use substitution to integrate $x \sqrt{2 x - 1}$ $xsqrt(2x-1)$ ?

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How do you use substitution to integrate $x \sqrt{2 x - 1}$ $xsqrt(2x-1)$ ?