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Help With 2 Calculus Questions?

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Integrate below functions

asked Apr 9, 2013 in CALCULUS by andrew Scholar

5 Answers

0 votes

24.

integral h open parentheses x close parentheses d x equal integral fraction numerator open parentheses x to the power of 2 minus 4 close parentheses over denominator x to the power of 3 plus 2 x to the power of 2 plus x plus 2 end fraction

integral h open parentheses x close parentheses d x equal integral fraction numerator open parentheses x minus 2 close parentheses open parentheses x plus 2 close parentheses over denominator x to the power of 2 open parentheses x plus 2 close parentheses plus x plus 2 end fraction      (Simplify)

integral h open parentheses x close parentheses d x equal integral fraction numerator open parentheses x minus 2 close parentheses open parentheses x plus 2 close parentheses over denominator open parentheses x plus 2 close parentheses open parentheses x to the power of 2 plus 1 close parentheses end fraction         (Take out common term  x + 2)

integral h open parentheses x close parentheses d x equal integral fraction numerator open parentheses x minus 2 close parentheses over denominator open parentheses x to the power of 2 plus 1 close parentheses end fraction                     (Cancel the term x + 2)

integral h open parentheses x close parentheses d x equal integral fraction numerator open parentheses x close parentheses over denominator open parentheses x to the power of 2 plus 1 close parentheses end fraction d x space minus integral fraction numerator 2 over denominator x to the power of 2 plus 1 end fraction d x

integral h open parentheses x close parentheses d x equal 1 over 2 integral fraction numerator 2 open parentheses x close parentheses over denominator open parentheses x to the power of 2 plus 1 close parentheses end fraction d x space minus 2 integral fraction numerator 1 over denominator x to the power of 2 plus 1 end fraction d x  (Integration formulas)

integral h open parentheses x close parentheses d x equal 1 over 2 italic log open vertical bar x to the power of 2 plus 1 close vertical bar space minus 2 italic tan to the power of minus 1 end exponent open parentheses x close parentheses space plus C

 

answered Apr 13, 2013 by diane Scholar
0 votes

30.

 

integral f open parentheses x close parentheses d x equal integral fraction numerator x to the power of 2 minus 6 x space minus 7 over denominator x plus 1 end fraction d x    

 integral f open parentheses x close parentheses d x equal integral fraction numerator x to the power of 2 minus 7 x plus x minus 7 over denominator x plus 1 end fraction d x

integral f open parentheses x close parentheses d x equal integral fraction numerator x open parentheses x minus 7 close parentheses plus 1 open parentheses x minus 7 close parentheses over denominator x plus 1 end fraction d x

integral f open parentheses x close parentheses d x equal integral fraction numerator open parentheses x minus 7 close parentheses open square brackets x plus 1 close square brackets over denominator x plus 1 end fraction d x                               (Take out common term x -7)

integral f open parentheses x close parentheses d x equal integral open parentheses x minus 7 close parentheses d x                                               (Cancel the term x + 1)

integral f open parentheses x close parentheses d x equal x to the power of 2 over 2 minus 7 x plus C                                             ( Apply Integration formulas)

 

answered Apr 13, 2013 by diane Scholar
0 votes

30.

 

integral f open parentheses x close parentheses d x equal integral fraction numerator x to the power of 2 minus 6 x space minus 7 over denominator x plus 1 end fraction d x    

 integral f open parentheses x close parentheses d x equal integral fraction numerator x to the power of 2 minus 7 x plus x minus 7 over denominator x plus 1 end fraction d x 

integral f open parentheses x close parentheses d x equal integral fraction numerator x open parentheses x minus 7 close parentheses plus 1 open parentheses x minus 7 close parentheses over denominator x plus 1 end fraction d x

integral f open parentheses x close parentheses d x equal integral fraction numerator open parentheses x minus 7 close parentheses open square brackets x plus 1 close square brackets over denominator x plus 1 end fraction d x                               (Take out common term x -7)

integral f open parentheses x close parentheses d x equal integral open parentheses x minus 7 close parentheses d x                                               (Cancel the term x + 1)

integral f open parentheses x close parentheses d x equal x to the power of 2 over 2 minus 7 x plus C                                             ( Apply Integration formulas)

 

 

integral f open parentheses x close parentheses d x equal integral fraction numerator x to the power of 2 minus 6 x space minus 7 over denominator x plus 1 end fraction d x    

 

 

                              

answered Apr 13, 2013 by diane Scholar
0 votes

24) The function is image

image

image

image

image

image

image

image

image

image

image

image

image

image

Substitute u value.

image

answered May 27, 2014 by joly Scholar
0 votes

30) The function is image.

image

image

image

image

image

image

image

image

image

answered May 27, 2014 by joly Scholar

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