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int(sqrt(x^2+10 x+24))/(x+5)dx is equal ...

`int(sqrt(x^2+10 x+24))/(x+5)dx` is equal to

A

`sqrt(x^(2)+10x+24)+sec^(-1)(x+5)+c`

B

`sqrt(x^(2)+10x+24)-"cosec"^(-1)(x+5)+c`

C

`sec^(-1)(x+5)-sqrt(x^(2)+10x+24)+c`

D

`sqrt(x^(2)+10x+24)-sec^(-1)(x+5)+c`

Text Solution

Verified by Experts

The correct Answer is:
D
` I=int(sqrt(x^(2)+10x+24))/(x+5)dx=int(sqrt((x+5)^(2)-1))/(x+5)dx`
`"Put " x+5=sec theta implies dx=sec theta tan theta d theta `
`:. I=int(tan theta)/(sec theta)*sec theta tan theta d theta `
`=int(sec^(2) theta-1)d theta`
`=tan theta-theta+C`
`=sqrt(x^(2)+10x+24)-sec^(-1)(x+5)+c`
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CENGAGE-INDEFINITE INTEGRATION-Exercise (Single)
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  2. int sqrt((x^(2)+1)/(x^(2)(1-x^(2))))dx=

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