Advertisements
Advertisements
Question
Evaluate the following:
`int sqrt(5 - 2x + x^2) "d"x`
Solution
Let I = `int sqrt(5 - 2x + x^2) "d"x`
= `int sqrt(x^2 - 2x + 5) "d"x`
= `int sqrt(x^2 - 2x + 1 - 1 + 5) "d"x` ....(Making perfect square)
= `int sqrt((x - 1)^2 + 4) "d"x`
= `int sqrt((x - 1)^2 + (2)^2) "d"x`
= `(x - 1)/2 sqrt((x - 1)^2 + (2)^2) + 4/2 log|(x - 1) + sqrt((x + 1)^2 + (2)^2)| + "C"` .......`[because int sqrt(x^2 + "a"^2) "d"x = x/2 sqrt(x^2 + "a"^2) + "a"^2/2 {log|x + sqrt(x^2 + "a"^2)|} + "C"]`
= `(x - 1)/2 sqrt(x^2 + 1 - 2x + 4) + 2log |(x - 1) + sqrt(x - 1) + sqrt(x^2 + 1 - 2x + 4)| + "C"`
= `(x - 1)/2 sqrt(x^2 - 2x + 5) + 2log|(x - 1) + sqrt(x^2 - 2x + 5)| + "C"`
Hence, I = `(x - 1)/2 sqrt(x^2 - 2x + 5) + 2log|(x - 1) + sqrt(x^2 - 2x + 5)| + "C"`
APPEARS IN
RELATED QUESTIONS
Integrate the following w.r.t. x `(x^3-3x+1)/sqrt(1-x^2)`
` ∫ cot^3 x "cosec"^2 x dx `
Evaluate the following integrals:
Evaluate the following integrals:
Evaluate the following integrals:
Evaluate the following integrals:
Evaluate the following integrals:
Evaluate the following integral:
Evaluate:\[\int\frac{\left( 1 + \log x \right)^2}{x} \text{ dx }\]
Evaluate:\[\int\frac{\log x}{x} \text{ dx }\]
Evaluate: \[\int 2^x \text{ dx }\]
Evaluate: \[\int\frac{x^3 - x^2 + x - 1}{x - 1} \text{ dx }\]
Evaluate:\[\int\frac{e\tan^{- 1} x}{1 + x^2} \text{ dx }\]
Evaluate: \[\int\frac{1}{x^2 + 16}\text{ dx }\]
Evaluate: \[\int\left( 1 - x \right)\sqrt{x}\text{ dx }\]
Evaluate: \[\int\frac{x + \cos6x}{3 x^2 + \sin6x}\text{ dx }\]
Evaluate:
\[\int \cos^{-1} \left(\sin x \right) \text{dx}\]
Evaluate the following:
`int sqrt(1 + x^2)/x^4 "d"x`
Evaluate the following:
`int ("d"x)/sqrt(16 - 9x^2)`
Evaluate the following:
`int x/(x^4 - 1) "d"x`
