If ∫(cosx+e2x+1x)dx=−asinx+e2xb+logx+c, then b+a=

∫(cosx+e2x+1x)dx=sinx+e2x2+logx+c∴a=−1, b=2∴b+a=2−1=1

∫(cosx+e2x+1x)dx=sinx+e2x2+logx+c∴a=−1, b=2∴b+a=2−1=1

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If $\int (\cos x + e^{2 x} + \frac{1}{x}) d x = - a \sin x + \frac{e^{2 x}}{b} + \log x + c$ , then $b + a =$

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$\int (\cos x + e^{2 x} + \frac{1}{x}) d x = \sin x + \frac{e^{2 x}}{2} + \log x + c$

$∴ a = - 1, b = 2$

$∴ b + a = 2 - 1 = 1$

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