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The integral inte^x(f(x)+f\'(x))dx can b...

The integral `inte^x(f(x)+f\'(x))dx` can be solved by using integration by parts such that: `I=inte^xf(x)dx+inte^xf\'(x)dx=e^xf(x)-inte^xf\'(x)dx+inte^xf\'(x)dx=e^xf(x)+C` , and `inte^(ax)(f(x)+(f\'(x))/a)dx=e^(ax)f(x)/a+C` ,Now answer the question:`int(e^x(2-x^2))/((1-x)sqrt(1-x^2))dx` (A) `e^xsqrt((1-x)/(1+x))+C` (B) `e^xsqrt((1+x)/(1-x))+C` (C) `e^xsqrt((2-x)/(2+x))+C` (D) none of these

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The integral inte^x(f(x)+f\'(x))dx can be solved by using integration by parts such that: I=inte^xf(x)dx+inte^xf\'(x)dx=e^xf(x)-inte^xf\'(x)dx+inte^xf\'(x)dx=e^xf(x)+C , and inte^(ax)(f(x)+(f\'(x))/a)dx=e^(ax)f(x)/a+C ,Now answer the question: inte^x x^x(2+logx)= (A) e^x x^xlogx+C (B) e^x+x^x+C (C) e^x x(logx)^2+C (D) e^x.x^x+C

The integral inte^x(f(x)+f\'(x))dx can be solved by using integration by parts such that: I=inte^xf(x)dx+inte^xf\'(x)dx=e^xf(x)-inte^xf\'(x)dx+inte^xf\'(x)dx=e^xf(x)+C , and inte^(ax)(f(x)+(f\'(x))/a)dx=e^(ax)f(x)/a+C ,Now answer the question: int{log_e(log_ex)+1/(log_ex)^2}dx is equal to (A) log_e(log_ex)+C (B) xlog_e(log_ex)-x/log_ex+C (C) x/log_ex-log_ex+C (D) log_e(log_ex)-x/log_ex+C

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int[f(x)+xf'(x)]dx=xf(x)+c

int[f(x)+xf'(x)]dx=

int[f(x)+x.f'(x)]dx=

If inte^(x)(1+x^(2))/((1+x)^(2))dx=e^(x)f(x)+c , then f(x)=

int e^xsqrt(e^(x)-1)dx