CBSE Guess > Papers > Important Questions > Class XII > 2010 > Maths > Mathematics By Mr. M.P.Keshari

Integrals

Q.4. Evaluate : ∫[(x² + x + 1)/{(x + 2)(x + 1)²}] dx.

Solution :

Do yourself. [Ans. = 3log|x + 2| – 2log|x + 1| – 1/(x + 1) + c.]

7.6. Integration by Parts.

Q.1. Evaluate : ∫x sin –1x dx.

Solution :

Let I = ∫x sin –1x dx.
Using integration by parts, taking x as second function we get
I = sin –1x.(x²)/2 – ∫[1/{√(1 – x²)}.(x²)/2] dx + c
= x²/2 sin –1x – 1/2∫[x²/√(1 – x²)] dx + c
= x²/2 sin –1x + 1/2∫[(1 – x² – 1)/{√(1 – x²)}] dx
= x²/2 sin –1x + 1/2∫[(1 – x²)/{√(1 – x²)}] dx – 1/2∫dx/√(1 – x²) + c
= 1/2[x² sin –1x + ∫√(1 – x²)dx – ∫dx/√(1 – x²)dx + c
= 1/2[x² sin –1x + x/2√(1 – x²)dx + 1/2sin –1x – sin –1x] + c
= 1/2[x² sin –1x + x/2√(1 – x²)dx – 1/2sin –1x] + c
= x²/2 sin –1x + x/4√(1 – x²) – 1/4sin –1x + c. [Ans.]

Q.2. Evaluate : ∫x² cot –1 x dx.

Solution :

Let I = ∫x² cot –1 x dx
Integrating by parts taking x² as second function, we get
I = (x³/3) cot –1 x – ∫(x³/3).{(–1)/(x² + 1)}dx
= (1/3) x³cot –1 x + (1/3)∫{x3/(x² + 1) }dx
= (1/3) x³ cot –1 x +(1/3) ∫[x – x/(x² + 1)]dx
= (1/3) x³ cot –1 x + (1/3)[x2/2 – 1/2 log|x² + 1|] + c. [Ans.]

Q.3. Evaluate : ∫x.log(x + 1)dx.

Solution :

Let I = ∫x.log(x + 1)dx
= log(x + 1)∫xdx – ∫[d/dx{log(x + 1)}∫.xdx].dx [Integrating by parts]
= log(x + 1).(x²/2) – ∫[1/(x + 1)].(x²/2)dx
= x²/2log(x + 1) – 1/2∫[x²/(x + 1)]dx
= x²/2log(x + 1) – 1/2∫[x – 1 + 1/(x + 1)]dx
= x²/2log(x + 1) – 1/2[x² /2 – x + log|x + 1| ] + c
= x²/2log(x + 1) – x²/4 – x/2 – 1/2log(x + 1) + c
= 1/2(x² – 1)log(x + 1) – x²/4 – x/2 + c. [Ans.]

Paper By Mr. M.P.Keshari
Email Id : [email protected]
Ph No. : 09434150289