Indefinite Integration - with trignometric substitution ∫ x − 1 x + 1 d xThe tip I got was to use x = sec 2 ⁡ ( a ). I don't no how to continue

Question

Indefinite Integration - with trignometric substitution  ∫                              x                −        1                              x                +        1                  d  xThe tip I got was to use   x  =      sec    2    ⁡  (  a  ). I don&#039;t no how to continue

Dahn2cm1p · Accepted Answer

x                            =                  sec          2                ⁡        t                            d        x                            =        2                  sec          2                ⁡        t        tan        ⁡        t                d        t                                                                                x                            −              1                                                      x                            +              1                                                          =                                            (                              x                            −              1              )              (                              x                            −              1              )                                      (                              x                            +              1              )              (                              x                            −              1              )                                                                        =                                            x                        −            1                                x            −            1                                                            =                              sec            ⁡            t            −            1                                tan            ⁡            t                                              I                            =        ∫                                                            x                            −              1                                                      x                            +              1                                              d        x                                          =        ∫        2                  sec          2                ⁡        t        (        sec        ⁡        t        −        1        )                d        t                                          =        2        ∫                  sec          3                ⁡        t                d        t        −        2        ∫                  sec          2                ⁡        t                d        t                                          =        2        ∫                  sec          3                ⁡        t                d        t        −        2        tan        ⁡        t                            ∫                  sec          3                ⁡        t                d        t                            =        ∫        sec        ⁡        t                d        (        tan        ⁡        t        )                                          =        sec        ⁡        t        tan        ⁡        t        −        ∫        tan        ⁡        t                d        (        sec        ⁡        t        )                                          =        sec        ⁡        t        tan        ⁡        t        −        ∫                  tan          2                ⁡        t        sec        ⁡        t                d        t                                          =        sec        ⁡        t        tan        ⁡        t        −        ∫        (                  sec          2                ⁡        t        −        1        )        sec        ⁡        t                d        t                                          =        sec        ⁡        t        tan        ⁡        t        −        ∫                  sec          3                ⁡        t                d        t        +        ∫        sec        ⁡        t                d        t                            2        ∫                  sec          3                ⁡        t                d        t                            =        sec        ⁡        t        tan        ⁡        t        +        ∫        sec        ⁡        t                d        t                            ∫        sec        ⁡        t                d        t                            =        ∫                              sec            ⁡            t            tan            ⁡            t            +                          sec              2                        ⁡            t                                sec            ⁡            t            +            tan            ⁡            t                                  d        t                                          =        ∫                              d            (            sec            ⁡            t            +            tan            ⁡            t            )                                sec            ⁡            t            +            tan            ⁡            t                                                            =        ln        ⁡                  |                sec        ⁡        t        +        tan        ⁡        t                  |                +        C                            I                            =        (        sec        ⁡        t        −        2        )        tan        ⁡        t        +        ln        ⁡                  |                sec        ⁡        t        +        tan        ⁡        t                  |                +        C                                          =        (                  x                −        2        )                  x          −          1                +        ln        ⁡        (                  x                +                  x          −          1                )        +        C

Indefinite Integration - with trignometric substitution ∫ <msqrt>

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