Integrate: ∫ x 2 − x + 1 d x

Question

Integrate:   ∫                    x        2            −      x      +      1        d  x

Alexis Fields · Accepted Answer

I                            =        ∫                              x                          2                                −          x          +          1                d        x                                          =        x                              x                          2                                −          x          +          1                −        ∫        x                              2            x            −            1                                2                                          x                                  2                                            −              x              +              1                                      d        x                                          =        x                              x                          2                                −          x          +          1                −        ∫                              2                          (                              x                                  2                                            −              x              +              1              )                        +                          (              x              −                              1                2                            )                        −                          3              2                                            2                                          x                                  2                                            −              x              +              1                                      d        x                                          =        x                              x                          2                                −          x          +          1                −        I        −                  1          4                ∫                              2            x            −            1                                              x                              2                                      −            x            +            1                          d        x        +                  3          4                ∫                              d            x                                              x                              2                                      −            x            +            1                                                            =                  1          2                          (          x                                    x                              2                                      −            x            +            1                    −                      1            2                                              x                              2                                      −            x            +            1                    )                +                  3          8                ∫                              d            x                                                              (                x                −                                  1                  2                                )                                            2                                      +                          3              4                                          For the last integral, letting   x  −      1    2    =            3        2    tan  ⁡  θ yields                    ∫                              d            x                                                              (                x                −                                  1                  2                                )                                            2                                      +                          3              4                                                          =        ∫                                                            3                            2                                      sec                              2                                      ⁡            θ            d            θ                                              3              4                                      tan                              2                                      ⁡            θ            +                          3              4                                                                        =        ∫        sec        ⁡        θ        d        θ                                          =        ln        ⁡                  |                sec        ⁡        θ        +        tan        ⁡        θ                  |                +                  C          1                                                  =        ln        ⁡                  |                                    2                                                x                                      2                                                  −                x                +                1                                                    3                                +                                    2              x              −              1                                      3                                |                +                  C          1                                                  =        ln        ⁡                  |          2          x          −          1          +          2                                    x                              2                                      −            x            +            1                    |                +                  C          2                    Now we can conclude that  I  =            2      x      −      1        4              x              2              −    x    +    1    +      3    8    ln  ⁡      |    2    x    −    1    +    2                  x                  2                    −      x      +      1        |    +  C

Lillianna Andersen · Accepted Answer

Alternate Method (Using hyperbolic functions)Letting   x  −      1    2    =            3        2    sinh  ⁡  θ yields                    ∫                              d            x                                                              (                x                −                                  1                  2                                )                                            2                                      +                          3              4                                                          =        ∫                                                            3                            2                        cosh            ⁡            θ            d            θ                                              3              4                                      sinh                              2                                      ⁡            θ            +                          3              4                                                                        =        ∫        d        θ                                          =        θ                                          =                  sinh                      −            1                          ⁡                  (                                    2              x              −              1                                      3                                )                                                                                                  I                =                                                      2                    x                    −                    1                                    4                                                                      x                                          2                                                        −                  x                  +                  1                                +                                  3                  8                                                  sinh                                      −                    1                                                  ⁡                                  (                                                            2                      x                      −                      1                                                              3                                                        )                                +                C

Integrate: ∫ <mrow class="MJX-TeXAtom-ORD"> <msqrt> x

Answered question

Answer & Explanation

New Questions in Calculus 1