How does one go from this: ∫ l 0 ( 1 − ξ / x 1 ) κ 0 2 ( 1 − x 0 / x 1 ) 2 − l 0 2 ( 1 − ξ / x 1 ) 2 d ξ

Question

How does one go from this:   ∫                    l        0            (      1      −              ξ                    /                    x        1            )                                κ                0        2            (      1      −              x        0                    /                    x        1                    )        2            −              l        0        2            (      1      −              ξ                    /                    x        1                    )        2              d      ξ

Jake Mcpherson · Accepted Answer

We then make the substitution   u  =      l    0    (  1  −  ξ      /        x    0    ) which also gives   −            x      0              l      0            d    u  =      d    ξ and allowing us to transform the integral into            x      0              l      0        ∫            −      u                      a        2            −              u        2                  d    uFor this integral, we need to use inspection. If we consider the             a      2        −          u      2      , the derivative of this is                                          d                                              d                        u                                                a            2                    −                      u            2                                              =                  1          2                          1                                    a              2                        −                          u              2                                                            d                                              d                        u                                    (                      a            2                    −                      u            2                    )                                                  =                  1          2                          1                                    a              2                        −                          u              2                                                (          −          2          u          )                                                  =                              −            u                                              a              2                        −                          u              2                                          This happens to be our integrand, so we can now calculate the integral as follows                                          x            0                                l            0                          ∫                              −            u                                              a              2                        −                          u              2                                                d                u                            =                              x            0                                l            0                          ∫                              d                                              d                        u                                                a            2                    −                      u            2                                    d                u                                          =                              x            0                                l            0                                    [                                    a              2                        −                          u              2                                +          c          ]

How does one go from this: ∫ l 0 </msub>

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