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Integrals

7.4. Integration of Some Particular Function.

Q.1. Evaluate : ∫√[(1 – x)/(1 + x)] dx.

Solution :

We have, ∫√[(1 – x)/(1 + x)] dx = ∫√[{(1 – x)(1 – x)}/{(1 + x)(1 – x)}] dx
= ∫[(1 – x)/√(1 – x²)] dx
= ∫[1/√(1 – x²)] dx + 1/2 ∫[(– 2x)/√(1 – x²)] dx
= sin –1x + 1/2 ∫dt/√t []
= sin –1x + 1/2 ∫t–1/2dt
= sin –1x + 1/2 [(t1/2)/(1/2)] + c
= sin –1x + √(1 – x²) + c. [Ans.]

Q.2. Evaluate : ∫dx/√(x² – 3x + 2).

Solution :

We have,
∫dx/√(x² – 3x + 2) = ∫dx/√{(x² – 3x + 9/4) – 1/4}
= ∫dx/√{(x – 3/2)² – (1/2)²}
= [(x – 3/2)√{(x – 3/2)² – (1/2)2}]/2 – {(1/2)2}/2 log |(x – 3/2) + √{(x – 3/2)² – (1/2)²}| + c
= [(x – 3/2)√(x² – 3x +2)]/2 – 1/8 log |(x – 3/2) + √(x² – 3x + 2)| + c. [Ans.]

Q.3. Evaluate : ∫[(3x + 1)/(2x² – 2x + 3)] dx.

Solution :

We have
I = ∫[(3x + 1)/(2x² – 2x + 3)] dx
Here, 3x + 1 = 3/4.d/dx(2x² – 2x + 3) + 5/2.
Therefore, I = ∫[{3/4(4x – 2) + 5/2}/(2x² – 2x + 3)] dx
= 3/4∫[(4x – 2)/(2x² – 2x + 3)] dx + 5/2∫dx/(2x² – 2x + 3)
= 3/4I1 + 5/2I² + c ----------------- (1)
Where, I1 = ∫[(4x – 2)/(2x² – 2x + 3)] dx
Put 2x² – 2x + 3 = t, then (4x – 2) dx = dt
Therefore, I1 = ∫dt/t = log |t| = log |2x² – 2x + 3| ---------------- (2)
and I2 = ∫dx/(2x² – 2x + 3)
= 1/2∫dx/(x² – x + 3/2)
= 1/2∫dx/[(x – 1/2)² + (√5/2)²]
= 1/2.1/(√5/2) tan –1[(x – 1/2)/(√5/2)]
[As,∫dx/(x² + a²) = 1/a tan –1(x/a)]
= 1/√5tan –1[(2x – 1)/√5] ----------------------- (3)
using (1), (2) and (3), we get
I = 3/4log|2x² – 2x + 3| + √5/2tan –1[(2x – 1)/√5] + c. [Ans.]

Paper By Mr. M.P.Keshari
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