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प्रश्न
Evaluate the following integral:
`int_0^3 (x dx)/(sqrt(x + 1)+ sqrt(5x + 1))`
उत्तर
`x/(sqrt(x + 1) + sqrt(5x + 1)) = x/(sqrt(x + 1) + sqrt(5 + 1)) xx (sqrt(x + 1) - sqrt(5x + 1))/(sqrt(x + 1) - sqrt(5x + 1)`
= `(x[sqrt(x + 1) - sqrt5x + 1])/((x + 1) - (5x + 1))`
= `(x(sqrt(x + 1) - sqrt(5x + 1)))/((x + 1) - (5x + 1))`
= `- 1/4 (sqrt(x + 1) - sqrt(5x + 1))`
So integral becomes
`- 1/4 int_0^3 (sqrt(x + 1) - sqrt(5x + 1)) "d"x = - 1/4 [(x + 1)^(3/2)/(3/2) - (5x + 1)^(3/2)/(3/2)^5]_0^3`
= `- 1/4 [4^(3/2)/(3/2) - (16)^(3/2)/(15/2)] + 1/4 [(1)^(3/2)/(3/2) - (-1)^(3/2)/(15/2)]`
= `- 1/4 (2/3 (8) - 2/15(64)) + 1/4[2/3 - 2/15]`
= `1/4 (16/3 - 128/15 - 2/3 + 2/15)`
= `- 1/4[14/3 - 126/15]`
= `- 1/4((70 - 126)/15)`
= `14/15`
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