InterviewSolution
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InterviewSolution
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Integral of `(x^(3)+3x+4)/(sqrtx)` isA. `(2)/(7)x^(5//2)+(2)/(3)x^(3//2)+8x^(1//2)+C`B. `(2)/(7)x^(7//2)+2x^(3//2)+8x^(1//2)+C`C. `(1)/(7)x^(7//2)+2x^(3//2)+8x^(1//2)+C`D. `x^(7//2)+3x^(3//2)+4x^(1//2)+C` |
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Answer» Correct Answer - B `int(x^(3)+3x+4)/(sqrtx)dx`. `int(x^(3)+3x+4)/(sqrtx)dx=int(x^(3))/(sqrtx)dx+3int(x)/(sqrtx)dx+4int(1)/(sqrtx)dx` `=int x^(3).x^(-1//2)dx+3 intx^(1//2)dx+4 int x^(-1//2)dx` `=int x^(5//2)dx+3int x^(1//2)dx+4 intx^(-1//2)dx` `=(x^((5//2)+1))/((5//2)+1)+(3x^((1//2)+1))/((1//2)+1)+(4x^((-1//2)+1))/((-1//2)+1)+C` `" "[because int x^(n)dx=(x^(n+1))/(n+1)]` `=(x^(7//2))/(7//2)+(3x^(3//2))/(3//2)+(4x^(1//2))/(1//2)+C` `=(2)/(7)x^(7//2)+2x^(3//2)+8x^(1//2)+C` |
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