Indefinite Integration 1 Question 51

52. The value(s) of 01x4(1x)41+x2dx is (are)

(2010)

(a) 227π

(b) 2105

(c) 0

(d) 71153π2

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Solution:

  1. Let I=01x4(1x)41+x2dx=01(x41)(1x)4+(1x)4(1+x2)dx

=01(x21)(1x)4dx+01(1+x22x)2(1+x2)dx=01(x21)(1x)4+(1+x2)4x+4x2(1+x2)dx=01(x21)(1x)4+(1+x2)4x+441+x2dx=01x64x5+5x44x2+441+x2dx=x774x66+5x554x33+4x4tan1x=1746+5543+44π40=227π



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