Advertisements
Advertisements
प्रश्न
Evaluate the following:
`int sqrt(5 - 2x + x^2) "d"x`
उत्तर
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
संबंधित प्रश्न
Evaluate the following integrals:
Evaluate the following integral :-
Evaluate the following integral:
Evaluate:\[\int\frac{x^2}{1 + x^3} \text{ dx }\] .
Evaluate:\[\int\frac{\sec^2 \sqrt{x}}{\sqrt{x}} \text{ dx }\]
Evaluate: \[\int\frac{2}{1 - \cos2x}\text{ dx }\]
Evaluate : \[\int\frac{1}{x(1 + \log x)} \text{ dx}\]
Evaluate: `int_ (x + sin x)/(1 + cos x ) dx`
Evaluate the following:
`int ("d"x)/sqrt(16 - 9x^2)`
Evaluate the following:
`int sqrt(x)/(sqrt("a"^3 - x^3)) "d"x`
