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B
esin−1x+c
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C
1√1−x2esin−1x+c
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D
x2esin1X+c
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Solution
The correct option is Axesin−1x+c I=∫esin−1x[1+x√1−x2]dx Put sin−1x=t ⇒x=sint ⇒dx=costdt So, I=∫et(sint+cost)dt ⇒I=(sint+cost)et−∫(cost−sint)etdt ⇒I=(sint+cost)et−(cost−sint)et−∫et(sint+cost)dt ⇒2I=2sintet+C ⇒I=xesin−1x+C