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a
xtan−1x−x+12tan−1x+C
b
xtan−1x+C
c
−x+xtan−1x+C
d
None of these above I1=∫xx+1dx
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Correct option is D
I=∫1⋅tan−1xdx=tan−1x⋅x−∫11+x⋅12x⋅xdx=xtan−1x−12∫x1+xdxLet I1=∫xx+1dxAssume that x=t⇒ 12xdx=dt⇒dx=2tdt∴ I1=∫t⋅2t1+t2dt=2∫1+t21+t2−11+t2dt =2t−2tan−1t =2x−2tan−1x∴ I=xtan−1x−122x−2tan−1x+C =xtan−1x−x+tan−1x+C =(x+1)tan−1x−x+CSimilar Questions
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