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Evaluate the following integrals
$integral subscript negative 1 end subscript superscript 1 space straight x cubed space left parenthesis straight x to the power of 4 plus 1 right parenthesis cubed space dx$

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102.

Evaluate the following integrals
$integral subscript 0 superscript 1 straight x square root of 1 minus straight x squared end root dx$

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103.

Evaluate the following integrals:
$integral subscript negative 1 end subscript superscript 1 fraction numerator 5 straight x over denominator left parenthesis 4 plus straight x squared right parenthesis squared end fraction dx$

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104.

Evaluate the following integral:
$integral subscript 0 superscript 1 fraction numerator 5 straight x over denominator left parenthesis 4 plus straight x squared right parenthesis squared end fraction dx$

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Long Answer Type

105.
Evaluate the following integral:
$integral subscript 0 superscript 2 fraction numerator dx over denominator straight x plus 4 minus straight x squared end fraction$

Let I = $integral subscript 0 superscript 2 fraction numerator dx over denominator straight x plus 4 minus straight x squared end fraction space equals space integral subscript 0 superscript 2 fraction numerator dx over denominator 4 minus left parenthesis straight x squared minus straight x right parenthesis end fraction$ .
$equals space integral subscript 0 superscript 2 fraction numerator dx over denominator straight x plus 4 minus straight x squared end fraction space equals space integral subscript 0 superscript 2 fraction numerator dx over denominator 4 minus left parenthesis straight x squared minus straight x right parenthesis end fraction equals space integral subscript 0 superscript 2 fraction numerator dx over denominator open parentheses 4 plus begin display style 1 fourth end style close parentheses minus open parentheses straight x squared minus straight x plus begin display style 1 fourth end style close parentheses end fraction space equals space integral subscript 0 superscript 2 fraction numerator dx over denominator open parentheses begin display style fraction numerator square root of 17 over denominator 2 end fraction end style close parentheses squared minus open parentheses straight x minus begin display style 1 half end style close parentheses squared end fraction equals space fraction numerator 1 over denominator 2 cross times begin display style fraction numerator square root of 17 over denominator 2 end fraction end style end fraction open square brackets log space open vertical bar fraction numerator begin display style fraction numerator square root of 17 over denominator 2 end fraction end style plus open parentheses straight x minus begin display style 1 half end style close parentheses over denominator begin display style fraction numerator square root of 17 over denominator 2 end fraction end style minus open parentheses straight x minus begin display style 1 half end style close parentheses end fraction close vertical bar close square brackets subscript 0 superscript 2$

$equals fraction numerator 1 over denominator square root of 17 end fraction open square brackets log space open curly brackets fraction numerator begin display style fraction numerator square root of 17 over denominator 2 end fraction end style plus open parentheses 2 minus begin display style 1 half end style close parentheses over denominator begin display style fraction numerator square root of 17 over denominator 2 end fraction end style minus open parentheses 2 minus begin display style 1 half end style close parentheses end fraction close curly brackets minus log open curly brackets fraction numerator begin display style fraction numerator square root of 17 over denominator 2 end fraction end style minus begin display style 1 half end style over denominator begin display style fraction numerator square root of 17 over denominator 2 end fraction end style plus begin display style 1 half end style end fraction close curly brackets close square brackets$

$equals fraction numerator 1 over denominator square root of 17 end fraction open square brackets log space fraction numerator square root of 17 plus 3 over denominator square root of 17 minus 3 end fraction minus log fraction numerator square root of 17 minus 1 over denominator square root of 17 plus 1 end fraction close square brackets equals space fraction numerator 1 over denominator square root of 17 end fraction log open square brackets fraction numerator square root of 17 plus 3 over denominator square root of 17 minus 3 end fraction cross times fraction numerator square root of 17 plus 1 over denominator square root of 17 minus 1 end fraction close square brackets space equals space fraction numerator 1 over denominator square root of 17 end fraction log open parentheses fraction numerator 17 plus 3 plus 4 square root of 17 over denominator 17 plus 3 minus 4 square root of 17 end fraction close parentheses equals space fraction numerator 1 over denominator square root of 17 end fraction log open parentheses fraction numerator 5 plus square root of 17 over denominator 5 minus square root of 17 end fraction close parentheses space equals space fraction numerator 1 over denominator square root of 17 end fraction log open parentheses fraction numerator 5 plus square root of 17 over denominator 5 minus square root of 17 end fraction cross times fraction numerator 5 plus square root of 17 over denominator 5 plus square root of 17 end fraction close parentheses equals space fraction numerator 1 over denominator square root of 17 end fraction log open parentheses fraction numerator 25 plus 17 plus 10 square root of 17 over denominator 25 minus 17 end fraction close parentheses equals space fraction numerator 1 over denominator square root of 17 end fraction log open parentheses fraction numerator 42 plus 10 square root of 17 over denominator 8 end fraction close parentheses equals space fraction numerator 1 over denominator square root of 17 end fraction log open parentheses fraction numerator 21 plus 5 square root of 7 over denominator 4 end fraction close parentheses$

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Short Answer Type

106. Evaluate the following integral:

integral subscript negative 1 end subscript superscript 1 fraction numerator dx over denominator straight x squared plus 2 straight x plus 5 end fraction

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107.

Evaluate $integral subscript 0 superscript straight pi over 2 end superscript space square root of sin space straight ϕ end root space cos to the power of 5 straight ϕ space dϕ space$

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Long Answer Type

108.

Evaluate $integral subscript 0 superscript 2 fraction numerator 5 straight x plus 1 over denominator straight x squared plus 4 end fraction dx.$

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109.

Prove that: $integral subscript negative straight a end subscript superscript straight a square root of fraction numerator straight a minus straight x over denominator straight a plus straight x end fraction end root space dx space equals space straight a space straight pi.$

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110.

Evaluate $integral subscript 0 superscript straight pi over 2 end superscript fraction numerator sin space straight x space plus space cos space straight x over denominator square root of sinx space cosx end root end fraction dx.$