How do you integrate $\int \frac{x^{2} - 1}{\sqrt{2 x - 1}} d x$ $int (x^2-1)/sqrt(2x-1)dx$ ?

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How do you integrate $\int \frac{x^{2} - 1}{\sqrt{2 x - 1}} d x$ $int (x^2-1)/sqrt(2x-1)dx$ ?

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Answer 1 · 2017-01-06T15:17:12

The answer is $\frac{1}{20} {(2 x - 1)}^{\frac{5}{2}} + \frac{1}{6} {(2 x - 1)}^{\frac{3}{2}} - \frac{3}{4} {(2 x - 1)}^{\frac{1}{2}} + C$ $1/20(2x - 1)^(5/2) + 1/6(2x - 1)^(3/2) - 3/4(2x - 1)^(1/2) + C$

Explanation:

Let $u = 2 x - 1$ $u = 2x - 1$ . Then $d u = 2 d x$ $du = 2dx$ and $d x = \frac{1}{2} d u$ $dx = 1/2du$ .

$\int \frac{x^{2} - 1}{\sqrt{u}} \cdot \frac{1}{2} d u$ $int(x^2 - 1)/sqrt(u) * 1/2du$

We want to get rid of the x's. $x = \frac{u + 1}{2}$ $x = (u + 1)/2$ , so $x^{2} = {(\frac{u + 1}{2})}^{2}$ $x^2 = ((u + 1)/2)^2$

$\int \frac{{(\frac{u + 1}{2})}^{2} - 1}{\sqrt{u}} \cdot \frac{1}{2} d u$ $int(((u + 1)/2)^2 - 1)/sqrt(u) * 1/2du$

Expand:

$\frac{1}{2} \int \frac{(\frac{u^{2} + 2 u + 1}{4}) - 1}{\sqrt{u}} d u$ $1/2int(((u^2 + 2u + 1)/4) - 1)/sqrt(u)du$

$\frac{1}{2} \int \frac{u^{2} + 2 u + 1 - 4}{4 \sqrt{u}} d u$ $1/2int(u^2 + 2u + 1 - 4)/(4sqrt(u))du$

$\frac{1}{8} \int \frac{u^{2} + 2 u - 3}{\sqrt{u}}$ $1/8int(u^2 + 2u - 3)/sqrt(u)$

$\frac{1}{8} \int (\frac{u^{2}}{\sqrt{u}} + \frac{2 u}{\sqrt{u}} - \frac{3}{\sqrt{u}}) d u$ $1/8int(u^2/sqrt(u) + (2u)/sqrt(u) - 3/sqrt(u))du$

$\frac{1}{8} \int (u^{\frac{3}{2}} + 2 u^{\frac{1}{2}} - 3 u^{- \frac{1}{2}}) d u$ $1/8int(u^(3/2) + 2u^(1/2) - 3u^(-1/2))du$

$\frac{1}{8} (\frac{2}{5} u^{\frac{5}{2}} + \frac{4}{3} u^{\frac{3}{2}} - 6 u^{\frac{1}{2}})$ $1/8(2/5u^(5/2) + 4/3u^(3/2) - 6u^(1/2))$

$\frac{1}{20} u^{\frac{5}{2}} + \frac{1}{6} u^{\frac{3}{2}} - \frac{3}{4} u^{\frac{1}{2}}$ $1/20u^(5/2) + 1/6u^(3/2) - 3/4u^(1/2)$

$\frac{1}{20} {(2 x - 1)}^{\frac{5}{2}} + \frac{1}{6} {(2 x - 1)}^{\frac{3}{2}} - \frac{3}{4} {(2 x - 1)}^{\frac{1}{2}} + C$ $1/20(2x - 1)^(5/2) + 1/6(2x - 1)^(3/2) - 3/4(2x - 1)^(1/2) + C$

Hopefully this helps!

Answer 2 · 2017-11-15T18:08:19

$\int \frac{x^{2} - 1}{\sqrt{2 x - 1}} \cdot d x$ $int (x^2-1)/sqrt(2x-1)*dx$

= $\frac{1}{4} \cdot \int \frac{4 x^{2} - 4}{\sqrt{2 x - 1}} \cdot d x$ $1/4*int (4x^2-4)/sqrt(2x-1)*dx$

= $\frac{1}{4} \cdot \int \frac{{(2 x)}^{2} - 4}{\sqrt{2 x - 1}} \cdot d x$ $1/4*int ((2x)^2-4)/sqrt(2x-1)*dx$

After using $u = \sqrt{2 x - 1}$ $u=sqrt(2x-1)$ , $2 x = u^{2} + 1$ $2x=u^2+1$ , $2 d x = 2 u \cdot d u$ $2dx=2u*du$ or $d x = u \cdot d u$ $dx=u*du$ transforms, this integral became

$\frac{1}{4} \cdot \int \frac{{(2 x)}^{2} - 4}{\sqrt{2 x - 1}} \cdot d x$ $1/4*int ((2x)^2-4)/sqrt(2x-1)*dx$

= $\frac{1}{4} \cdot \int \frac{{(u^{2} + 1)}^{2} - 4}{u} \cdot (u \cdot d u)$ $1/4*int ((u^2+1)^2-4)/u*(u*du)$

= $\frac{1}{4} \cdot \int (u^{4} + 2 u^{2} - 3) \cdot d u$ $1/4*int (u^4+2u^2-3)*du$

= $\frac{1}{4} \cdot (\frac{u^{5}}{5} + \frac{2 u^{3}}{3} - 3 u) + C$ $1/4*(u^5/5+(2u^3)/3-3u)+C$

= $\frac{u^{5}}{20} + \frac{u^{3}}{6} - \frac{3 u}{4} + C$ $u^5/20+u^3/6-(3u)/4+C$

= $\frac{1}{20} \cdot {(2 x - 1)}^{\frac{5}{2}} + \frac{1}{6} \cdot {(2 x - 1)}^{\frac{3}{2}} - \frac{3}{4} \cdot \sqrt{2 x - 1} + C$ $1/20*(2x-1)^(5/2)+1/6*(2x-1)^(3/2)-3/4*sqrt(2x-1)+C$

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How do you integrate $\int \frac{x^{2} - 1}{\sqrt{2 x - 1}} d x$ $int (x^2-1)/sqrt(2x-1)dx$ ?

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