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a
1a2+b2logtanπ4−x2+12tan−1ab+K
b
1a2+b2logtanπ4−x2−12tan−1ab+C
c
1a2+b2logtanπ4+x2−12tan−1ab+K
d
1a2+b2logtanπ4+x2+12tan−1ab+C
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Correct option is C
Let J=∫secxatanx+bdx=∫1asinx+bcosxdx=∫1a2+b2aa2+b2sinx+ba2+b2cosxdx Choose sinα=aa2+b2,cosα=ba2+b2,tanα=ab⇒α=tan−1abJ=1a2+b2∫1sinαsinx+cosαcosxdxJ=1a2+b2∫1cos(x−α)dx=1a2+b2∫sec(x−α)dxJ=1a2+b2logtanπ4+x−α2+CJ=1a2+b2logtanπ4+x2−12tan−1ab+KSimilar Questions
is equal to
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