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Find : ∫(x + 3)/(√(5 - 4x + x2)) dx

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Let x + 3 = A(2x - 4) + B

2A = 1 and B - 4A = 3

which gives A = 1/2 , B = 5

Given integral becomes

by (10 points)
but answer is wrong .answer of book is            - (5-4x+x^2)^1/2 +sin^-1(x+2/3) +c
by (10 points)
Question: $ (x+3)/[√(5 – 4x – x²)]

Let $ = Integral symbol, § = Angle theta

First of all, let's simplify the denominator.

5 – 4x – x² = 5 – (x² + 4x +   )

We complete the square, but we must be careful of the signs. Note that I put the x² term and the x term inside brackets with a minus sign outside. Naturally, the –4x becomes 4x inside the brackets. Now we can complete the square inside the brackets and 'add' on a similar amount outside the brackets; so nothing has changed.

We have,
5 – (x² – 4x + 2²) + 2² = 9 – (x + 2)²

Now that we're done simplifying, we put that back into the denominator.

$ (x+3)/(√(9 – (x+2)²))

If x + 2 = 3 sin §

Differentiating, we have dx = 3cos § d§

Let x + 3 = (x +2) + 1

= $ [(3sin§ + 1)/(√(9 – (x + 2)²))] • 3cos § d§

Remember (x + 2) = 3sin§

= $ [(3sin § + 1)(3cos §)]/(3cos §) • d§

3cos § cancels 3cos §

= $ (3sin § + 1)d§

= –3cos § + § + C

Drawing a right-angled triangle with sin § = (x + 2)/3 , cos § = {√[9 – (x+2)²]}

Substituting cos § and sin § into our final answer, we have

= {–3[√(9 – (x+2)²))]/ 3} + arcsin (x+2)/3 + C

Simplifying further,

= [–√(9–(x+2)²)]   +   arcsin(x+2)/3   +    C

= [–√(5 – 4x – x²) ] + arcsin (x+2)/3 + C


I hope this helps.

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