1.

EVALUATE.............

Answer»

Answer:

∫11+x4 dx

=∫1x21+x4x2 dx

=∫1x2x2+1x2 dx

=12∫2x2x2+1x2 dx

=12∫(1+1x2)−(1−1x2)x2+1x2 dx

=12∫(1+1x2)dxx2+1x2−12∫(1−1x2)dxx2+1x2

=12∫(1+1x2)dxx2+1x2−2+2−12∫(1−1x2)dxx2+1x2+2−2

=12∫d(x−1x)(x−1x)2+2−12∫d(x+1x)(x+1x)2−2

=12∫d(x−1x)(x−1x)2+(2√)2−12∫d(x+1x)(x+1x)2−(2√)2

★ Now, using STANDARD formula: ∫dxx2+a2=1atan−1(XA) & ∫dxx2−a2=12aln∣∣x−ax+a∣∣

=12⋅12√tan−1(x−1x2√)−12⋅122√ln∣∣∣x+1x−2√x+1x+2√∣∣∣+C

=122√tan−1(x2−1x2√)−142√ln∣∣x2−x2√+1x2+x2√+1∣∣+C


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