If ∫e^x√xx+√xdx=e^xAx+B√x+C+K, then A+B+C=

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MARKS Content Team · Accepted Answer

Correct answer: (B) 2. I=∫e^x√xx+√xdx Let √x=t 1/(2√x)dx=dt i.e. I=2∫e^tt^2+tdt =2∫t^2e^tdt+2∫te^tdt =2t^2e^t-2∫2te^tdt+2∫te^tdt =2t^2e^t-2∫te^tdt =2t^2e^t-2te^t-∫e^tdt=2t^2e^t-2te^t+2e^t+C^(') =e^x2x-2√x+2+C^(') Hence, A+B+C=2

If $\int \frac{e^{\sqrt{x}}}{\sqrt{x}} (x + \sqrt{x}) d x = e^{\sqrt{x}} [A x + B \sqrt{x} + C] + K$ , then $A + B + C =$

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If ∫exxx+xdx=exAx+Bx+C+K, then A+B+C=

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If $\int \frac{e^{\sqrt{x}}}{\sqrt{x}} (x + \sqrt{x}) d x = e^{\sqrt{x}} [A x + B \sqrt{x} + C] + K$ , then $A + B + C =$