How would one go about integrating the following? ∫ 0 5 x x − 2 d x

Question

How would one go about integrating the following?      ∫    0    5        x          x      −      2        d  x

Nigerkamg5 · Accepted Answer

The integrand       x          x      −      2       has a vertical asymptote at   x  =  2. Therefore the integral  I  =      ∫          0              5            x          x      −      2        d  x  =      ∫          0              2            x          x      −      2        d  x  +      ∫          2              5            x          x      −      2        d  x  =      I          1        +      I          2      is an improper integral of the second kind. It exists only if both integrals       I          1       and       I          2       exist. Let&#039;s compute the first one using the substitution   u  =  x  −  2  ,  d  x  =  d  u, as commented above                              I                      1                                              =                              ∫                      0                                2                                    x                      x            −            2                          d        x        =                  ∫                      −            2                                0                                                u            +            2                    u                d        u        =                  ∫                      −            2                                0                          1        d        u        +                  ∫                      −            2                                0                                    2          u                d        u                                          =                    2        +        2                  ∫                      −            2                                0                                    1          u                d        u        =        2        +        2        ln        ⁡                              |            u            |                                −            2                                0                          =        2        +        2                  lim                      ε            →            0                          ln        ⁡                              |            u            |                                −            2                                −            ε                          ,                ε        &amp;gt;        0                                          =                    2        +        2                  lim                      ε            →            0                          ln        ⁡                  |          −          ε          |                −        2        ln        ⁡                  |          −          2          |                                                  =                    2        −        2        ln        ⁡        2        +        2                  lim                      ε            →            0                          ln        ⁡        ε        .            Since       lim          ε      →      0        ln  ⁡  ε) doesn&#039;t exist, the integral       ∫          −      2              0            2    u    d  u does not converge and so does       I          1      . Similarly we might conclude that       I          2       does not converge as well.

How would one go about integrating the following? int_0^5 x/(x-2)dx

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