Use substitution to convert the integrals to integrals of rational functions. Then use partial fractions to evaluate the integrals. ∫ d x 1 + x + 1

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Use substitution to convert the integrals to integrals of rational functions. Then use partial fractions to evaluate the integrals.   ∫            d      x              1      +              x        +        1

incibracy5x · Accepted Answer

∫            d      x              1      +              x        +        1            Let       u          2        =  x  +  1  ⇒  2  u  d  u  =  d  xApply the substitution  ∫            d      x              1      +              x        +        1              =  ∫            2      u      d      u              1      +      u      By the long division            2      u              1      +      u        =  2  −      2          u      +      1      Therefore,  ∫            2      u      d      u              1      +      u        =  ∫  (  2  −      2          u      +      1        )  d  uIntegrate  ∫  (  2  −      2          u      +      1        )  d  u  =  2  u  −  2  ln  ⁡      |    u  +  1      |    +  CBack - substitute       u          2        =  x  +  1  ⇒  u  =      x    +    1    2  u  −  2  ln  ⁡      |    u  +  1      |    +  C  2      x    +    1    −  2  ln  ⁡      |        x    +    1    +  1      |    +  CResult:  2      x    +    1    −  2  ln  ⁡      |        x    +    1    +  1      |    +  C

Use substitution to convert the integrals to integrals of rational functions. Then use partial fractions to evaluate the integrals. int (dx)/(1+sqrt(x+1))

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