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.

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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.

Find The Antiderivative of more complex expressions. Need help finding the antiderivative of this o

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