∫1−1x2√x2+1+x+1dx = ?

Find ∫1−1x2√x2+1+x+1dx

2 Answers
Cem Sentin
Apr 3, 2018

∫1−1x2√x2+1+x+1⋅dx=13

Explanation:

∫1−1x2√x2+1+x+1⋅dx

=∫1−1x2⋅(x+1−√x2+1)⋅dx(x+1)2−(x2+1)

=∫1−1x2⋅(x+1−√x2+1)⋅dx2x

=12∫1−1x⋅(x+1−√x2+1)⋅dx

=12∫1−1(x2+x−x√x2+1)⋅dx

=[16x3+14x2−16(x2+1)32]1−1

=13

Cem Sentin
Apr 3, 2018

∫1−1x2√x2+1+x+1⋅dx=13

Explanation:

∫1−1x2√x2+1+x+1⋅dx

After using x=tany and dx=(secy)2⋅dy transforms, this integral became

∫π4−π4(tany)2secy+tany+1⋅(secy)2⋅dy

=∫π4−π4(tany)2⋅(tany+1−secy)⋅(secy)2⋅dy(tany+1)2−(secy)2

=∫π4−π4(tany)2⋅(tany+1−secy)⋅(secy)2⋅dy(tany)2+1+2tany−(secy)2

=∫π4−π4(tany)2⋅(tany+1−secy)⋅(secy)2⋅dy(secy)2+2tany−(secy)2

=∫π4−π4(tany)2⋅(tany+1−secy)⋅(secy)2⋅dy2tany

=12∫π4−π4tany⋅(tany+1−secy)⋅(secy)2⋅dy

=12∫π4−π4(tany)2⋅(secy)2⋅dy+12∫π4−π4(tany)⋅(secy)2⋅dy-12∫π4−π4(secy)3⋅tany⋅dy

=[16(tany)3+14(tany)2−16(secy)3]π4−π4

=13

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