Question

Evaluate the integral\(\int_{\rm{1}}^{\rm{2}} {\frac{{\sqrt {{{\rm{x}}^{\rm{2}}}{\rm{ - 1}}} }}{{\rm{x}}}} {\rm{dx}}\)

Short answer

Considering the integral \(\frac{{{\rm{32}}}}{{\rm{3}}}{\rm{ln(2) - }}\frac{{\rm{7}}}{{\rm{4}}}\)

Step-by-step solution

Both \({{\rm{x}}^{\rm{5}}}\)or\({\rm{In x}}\).

Both \({{\rm{x}}^{\rm{5}}}\) or \({\rm{In x}}\)would become simpler after differentiating for integration by parts, but integral would be a little more complicated (requiring integration by parts simply for itself), therefore we will chose integral\({\rm{u = Inx}}\).

\(\begin{aligned}{c}{\rm{u = lnx}}\;\\\;\;\;{\rm{du = }}\frac{{\rm{1}}}{{\rm{x}}}{\rm{dx}}\\{\rm{dv = }}{{\rm{x}}^{\rm{5}}}{\rm{dx}}\\\;{\rm{v = }}\frac{{\rm{1}}}{{\rm{6}}}{{\rm{x}}^{\rm{6}}}\;\;\;\;\;\end{aligned}\)

\(\begin{aligned}{c}_{\rm{1}}^{\rm{2}}{{\rm{x}}^{\rm{5}}}{\rm{lnxdx = uv - vdu}}\\{\rm{ = }}\frac{{\rm{1}}}{{\rm{6}}}{{\rm{x}}^{\rm{6}}}{\rm{lnx}}\;\;\;{\rm{ - }}\;\;\;\frac{{\rm{2}}}{{\rm{6}}}{{\rm{x}}^{\rm{6}}}{\rm{ \times }}\frac{{\rm{1}}}{{\rm{x}}}{\rm{dx}}\\{\rm{ = }}\frac{{\rm{1}}}{{\rm{6}}}{\rm{(64)ln2 - 0 - }}\frac{{\rm{1}}}{{\rm{6}}}_{\rm{1}}^{\rm{2}}{{\rm{x}}^{\rm{5}}}{\rm{dx}}\end{aligned}\)

Find\({\rm{u = Inx}}\).

\(\begin{aligned}{c}{\rm{ = }}\frac{{{\rm{32}}}}{{\rm{3}}}{\rm{ln(2) - }}\frac{{\rm{1}}}{{\rm{6}}}\frac{{\rm{1}}}{{\rm{6}}}{{\rm{x}}^{\rm{6}}}^{\rm{2}}\\{\rm{ = }}\frac{{{\rm{32}}}}{{\rm{3}}}{\rm{ln(2) - }}\frac{{\rm{1}}}{{\rm{6}}}\frac{{\rm{1}}}{{\rm{6}}}{\rm{(64) - }}\frac{{\rm{1}}}{{\rm{6}}}\\{\rm{ = }}\frac{{{\rm{32}}}}{{\rm{3}}}{\rm{ln(2) - }}\frac{{\rm{1}}}{{\rm{6}}}\frac{{{\rm{63}}}}{{\rm{6}}}\\{\rm{ = }}\frac{{{\rm{32}}}}{{\rm{3}}}{\rm{ln(2) - }}\frac{{\rm{1}}}{{\rm{6}}}\frac{{{\rm{21}}}}{{\rm{2}}}\\{\rm{ = }}\frac{{{\rm{32}}}}{{\rm{3}}}{\rm{ln(2) - }}\frac{{\rm{1}}}{{\rm{2}}}\frac{{\rm{7}}}{{\rm{2}}}\\{\rm{ = }}\frac{{{\rm{32}}}}{{\rm{3}}}{\rm{ln(2) - }}\frac{{\rm{7}}}{{\rm{4}}}{\rm{ = 5}}{\rm{.6436}}\end{aligned}\)

Integration is required for the final solution\(\frac{{{\rm{32}}}}{{\rm{3}}}{\rm{ln(2) - }}\frac{{\rm{7}}}{{\rm{4}}}\).