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

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

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

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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$\int x (\frac{\ln a^{a^{x / 2}}}{3 a^{5 x / 2} b^{3 x}} + \frac{\ln b^{b^{x}}}{2 a^{2 x} b^{4 x}}) d x (where a, b \in R^{+}) is equal to$

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$\frac{1}{6 \ln a^{2} b^{3}} \frac{1}{a^{2 x} b^{3 x}} \ln (a^{2 x} b^{3 x} e) + k$

$\frac{1}{6 \ln a^{2} b^{3}} \frac{1}{a^{2 x} b^{3 x}} \ln \frac{1}{e a^{2 x} b^{3 x}} + k$

$\frac{1}{6 \ln a^{2} b^{3}} a^{2 x} b^{3 x} \ln \frac{a^{2 x} b^{3 x}}{e} + k$

$- \frac{1}{6 \ln a^{2} b^{3}} \frac{1}{a^{2 x} b^{3 x}} \ln (a^{2 x} b^{3 x} e) + k$

answer is D.

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Detailed Solution

$I = \int x (\frac{\ln a^{a^{x / 2}}}{3 a^{5 x / 2} b^{3 x}} + \frac{\ln b^{b^{x}}}{2 a^{2 x} b^{4 x}}) d x = \int \frac{\ln a^{2 x} b^{3 x}}{6 a^{2 x} b^{3 x}} d x$

$Let a^{2 x} b^{3 x} = t . Then t \ln a^{2} b^{3} d x = d t . Therefore,$

$\begin{array}{l} I = \int \frac{1}{6 \ln a^{2} b^{3}} \frac{\ln t}{t^{2}} d t \\ = \frac{1}{6 \ln a^{2} b^{3}} (\frac{- \ln t}{t} - \int \frac{- 1}{t^{2}} d t) \\ = - \frac{1}{6 \ln a^{2} b^{3}} (\frac{\ln e t}{t}) + k \\ = - \frac{1}{6 \ln a^{2} b^{3}} (\frac{\ln a^{2 x} b^{3 x} e}{a^{2 x} b^{3 x}}) + k \end{array}$