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Evaluate the indefinite integral

0 votes

∫x^2*arctan(2x)dx?

asked May 27, 2013 in CALCULUS by angel12 Scholar

3 Answers

0 votes

integral x to the power of 2 cross times italic tan to the power of minus 1 end exponent open parentheses 2 x close parentheses d x equal x to the power of 2 integral italic tan to the power of minus 1 end exponent open parentheses 2 x close parentheses d x minus integral open square brackets 2 x integral italic tan to the power of minus 1 end exponent open parentheses 2 x close parentheses d x close square brackets d x

space space space space space space space space space space space space space space space space space equal x to the power of 2 open square brackets open parentheses 2 x italic tan to the power of minus 1 end exponent open parentheses 2 x close parentheses close parentheses minus 1 over 2 italic log open vertical bar 1 plus 4 x to the power of 2 close vertical bar plus C close square brackets minus integral 2 x open square brackets open parentheses 2 x italic tan to the power of minus 1 end exponent open parentheses 2 x close parentheses close parentheses minus 1 over 2 italic log open vertical bar 1 plus 4 x to the power of 2 close vertical bar plus C close square brackets d x                               (

space space space space space space space space space space space space space space space space space equal x to the power of 2 open square brackets open parentheses 2 x italic tan to the power of minus 1 end exponent open parentheses 2 x close parentheses close parentheses minus 1 over 2 italic log open vertical bar 1 plus 4 x to the power of 2 close vertical bar plus C close square brackets plus open square brackets minus 4 open parentheses integral x to the power of 2 italic tan to the power of minus 1 end exponent open parentheses 2 x close parentheses close parentheses d x plus x italic log open vertical bar 1 plus 4 x to the power of 2 close vertical bar plus C close square brackets d x   (Let y =  integral x to the power of 2 italic tan to the power of minus 1 end exponent open parentheses 2 x close parentheses d x )                    

space space space space space space space space space space space space space space space space space y equal x to the power of 2 open square brackets open parentheses 2 x italic tan to the power of minus 1 end exponent open parentheses 2 x close parentheses close parentheses minus 1 over 2 italic log open vertical bar 1 plus 4 x to the power of 2 close vertical bar plus C close square brackets minus 4 y plus integral open square brackets x italic log open vertical bar 1 plus 4 x to the power of 2 close vertical bar plus C close square brackets d x

space space space space space space space space space space space space space space space space space 5 y equal x to the power of 2 open square brackets open parentheses 2 x italic tan to the power of minus 1 end exponent open parentheses 2 x close parentheses close parentheses minus 1 over 2 italic log open vertical bar 1 plus 4 x to the power of 2 close vertical bar plus C close square brackets plus integral open square brackets x italic log open vertical bar 1 plus 4 x to the power of 2 close vertical bar close square brackets d x plus C x

space space space space space space space space space space space space space space space space space y equal x to the power of 2 over 5 open square brackets open parentheses 2 x italic tan to the power of minus 1 end exponent open parentheses 2 x close parentheses close parentheses minus 1 over 2 italic log open vertical bar 1 plus 4 x to the power of 2 close vertical bar plus C close square brackets plus 1 over 5 open square brackets integral open square brackets x italic log open vertical bar 1 plus 4 x to the power of 2 close vertical bar close square brackets d x close square brackets plus fraction numerator C x over denominator 5 end fraction.

 

answered Jun 5, 2013 by diane Scholar

Solution

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0 votes

Calculate the integral

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By using parts of integration image

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Now the integral becomes

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answered Jul 26, 2014 by david Expert
0 votes

  Contd...                      

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By division algorithm theorem

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answered Jul 26, 2014 by david Expert

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