Find The Antiderivative of more complex expressions.Need help finding the antiderivative of this one in particular d x 3 x ² + 1 .Would you kindly also give some tips on how to find the antiderivative of any expressions. I am having problems switching from differential calculus to integral calculus as finding the antiderivative is not as straight forward as finding the differential of an expression.

Question

Find The Antiderivative of more complex expressions.Need help finding the antiderivative of this one in particular             d      x              3      x              ²            +      1      .Would you kindly also give some tips on how to find the antiderivative of any expressions. I am having problems switching from differential calculus to integral calculus as finding the antiderivative is not as straight forward as finding the differential of an expression.

pampatsha · Accepted Answer

Step 1Generalize this. Start with the integral  I  =  ∫                    d            x                      a        2            +      (      b      x              )        2            Let   z  =                    b        x            a       so that       d    z  =            b      a              d    x. Thus                    I                            =                  a          b                ∫                                            d                        z                                              a              2                        +                          a              2                                      z              2                                                                        =                  1                      a            b                          ∫                                            d                        z                                1            +                          z              2                                          Step 2The integral is now easily seen to be   arctan  ⁡  z  +  C. Thus  ∫                    d            x                      a        2            +      (      b      x              )        2                  =          1              a        b              arctan    ⁡          (                        b          x                a            )        +    C  Now set   a  =  1 and   b  =      3  .

Holetaug · Accepted Answer

Step 1You need to know the formula  ∫      1                  x        2            +              a        2              d  x  =      1    a    arctan  ⁡      (          x      a        )    +  cSo your integral becomes:                    ∫                  1                      3                          x              2                        +            1                          d        x                            =                  1          3                ∫                  1                                    x              2                        +                          1              3                                      d        x        =                  1          3                ∫                  1                                    x              2                        +                                          (                                  1                                      3                                                  )                            2                                      d        x                                          =                  1          3                ×                  1                      1                          /                                      3                                      arctan        ⁡                  (                      x                          1                              /                                            3                                              )                +        c                                          =                  1                      3                          arctan        ⁡                  3                x        +        c            Step 2ANOTHER METHOD: By SubstitutionLet   x  =            1              3              tan    ⁡    θ  . So   d  x  =            1              3                    sec      2        ⁡    θ         d    θ  . Also   3      x    2    +  1  =  3      (          1      3              tan      2        ⁡    θ    )    +  1  =      tan    2    ⁡  θ  +  1  =      sec    2    ⁡  θ.Hence your integral becomes                    ∫                  1                      3                          x              2                        +            1                          d        x                            =        ∫                  1                                    sec              2                        ⁡            θ                                    (                      1                          3                                            sec            2                    ⁡          θ                     d          θ          )                                                  =        ∫                  1                      3                          d        θ                                          =                  1                      3                          θ        +        c                                                                                                  ⋯        (        i        )            Now since   x  =            1              3              tan    ⁡    θ  , so   θ  =  arctan  ⁡      3    x. Putting this in one we have the required answer as   ∫      1          3              x        2            +      1        d  x  =      1          3        arctan  ⁡      3    x  +  c.