Question
∫x2−2x3√x2−1dx is equal to
Answer: Option D
:
D
∫x2−2x3√x2−1dx=∫dxx√x2−1−2∫dxx3√x2−1=sec−1x−2∫secθtanθsec3θtanθdθ=[putx=secθ⇒dx=secθtanθdθ]=sec−1x−2∫cos2θdθ=sec−1x−2∫(1+cos2θ)dθ=sec−1x−(θ+sin2θ2)+C=sec−1x−sec−1x−√x2−1x2+C=−√x2−1x2+C
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:
D
∫x2−2x3√x2−1dx=∫dxx√x2−1−2∫dxx3√x2−1=sec−1x−2∫secθtanθsec3θtanθdθ=[putx=secθ⇒dx=secθtanθdθ]=sec−1x−2∫cos2θdθ=sec−1x−2∫(1+cos2θ)dθ=sec−1x−(θ+sin2θ2)+C=sec−1x−sec−1x−√x2−1x2+C=−√x2−1x2+C
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