integration of 1 by 1+x under root 3+2x-x2 dx

Dear student,

First let's write the term in the underroot :

-x^2+2x+3 = 4 - (x-1)^2

Now, put (x-1) = 2sin(theta).

so, dx = 2cos( theta). And 1+x will become : 2+2sin(theta)

Note, the under root term ( in denominator ) will become as : 4 - 4 sin^2(theta). Take out 4 from the underroot ( it will become 2) and the inside term of the root will be cos^2(theta), which will become cos(theta) and eventually root will be removed.

So, the integration will become 2cos(theta) d(theta) / { 2+2sin(theta)}×{2cos(theta)}

Now the 2 cos(theta) will get cancelled from the numerator and denominator and you will be left with only integration of d(theta) / 2(1+sin(theta))

Now, let's for a second just keep 2 aside from the denominator and multiplying the numerator and denominator by ( 1-sin(theta)). So, you will get 1-sin(theta) d(theta) / (1-sin^2(theta)).

Now, all the steps are now simple. The denominator term will become cos^2(theta).

And now, we have integration of ( 1 - sin(theta)) d(theta) / 2 cos^2(theta)

And now, just split the integration and you will get : {integration of d(theta) / 2cos^2(theta) } - {integration of sin(theta) d(theta) / 2cos^2(theta) }

After that you will get : { integration of sec^2(theta) d(theta) / 2 } - { integration of tan(theta) sec(theta) d(theta) /2 }.

So, the answer of first integration will be tan(theta) / 2 . And the answer of second integration will be sec(theta) /2 .

So, the combined result ( in terms of theta ) is = {tan(theta) - sec(theta) } / 2.

Now, tan(theta) in terms of x will be : (x-1) / √(4 - (x-1)^2) and sec( theta) in terms of x is : 2 / √(4 - (x-1)^2).

So, our final answer is = ((x-1) - 2) / 2×{3+2x-x^2} = (x-3)/2×{3+2x-x^2} ( if you have written the question correctly ).