Q.

∫xln⁡aax/23a5x/2b3x+ln⁡bbx2a2xb4xdx where a,b∈R+is equal to

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a

16ln⁡a2b3a2xb3xln⁡a2xb3xe+k

b

16ln⁡a2b31a2xb3xln⁡1ea2xb3x+k

c

16ln⁡a2b31a2xb3xln⁡a2xb3xe+k

d

−16ln⁡a2b31a2xb3xln⁡a2xb3xe+k

answer is D.

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Detailed Solution

I=∫xln⁡aax/23a5x/2b3x+ln⁡bbx2a2xb4xdx=∫ln⁡a2xb3x6a2xb3xdx Let a2xb3x=t. Then tln⁡a2b3dx=dt. Therefore, I=∫16ln⁡a2b3ln⁡tt2dt=16ln⁡a2b3−ln⁡tt−∫−1t2dt=−16ln⁡a2b3ln⁡ett+k=−16ln⁡a2b3ln⁡a2xb3xea2xb3x+k

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∫xln⁡aax/23a5x/2b3x+ln⁡bbx2a2xb4xdx where a,b∈R+is equal to