Class 12 RD Sharma Solutions – Chapter 19 Indefinite Integrals – Exercise 19.14

Question 1. Evaluate âˆ«1/ a2-b2x2 dx

Solution:

Let us assume I = âˆ« 1/ (a2-b2x2)dx

take 1/b2 common from above equation

= 1/b2 âˆ« 1/ (a2/b2-x2) dx

= 1/b2 âˆ« 1/ (a/b)2-x2) dx

Integrate the above eq then, we get

= 1/b2 1/ 2(a/b) log|(a/b)+x/ (a/b)-x| + c [since âˆ«1/ a2-x2 dx = 1/2a log|x+a/x-a| + c]

= 1/2ab log|a+bx/ a-bx| + c

Hence, I = 1/2ab log |a+bx/ a-bx| + c

Question 2. Evaluate âˆ« 1/ a2x2-b2 dx

Solution:

Let us assume I = âˆ« 1/ a2x2-b2 dx

take 1/a2 common from above equation

= 1/a2 âˆ« 1/ x2-(b2/a2) dx

= 1/a2 âˆ« 1/ x2-(b/a)2 dx

Integrate the above eq then, we get

= (1/a2) 1/(2b/a) log|x-(b/a)/x+(b/a)| + c [since âˆ«1/ x2-a2 dx = 1/2a log|x-a/x+a| + c]

= 1/2ab log|ax-b/ax+b| + c

Hence, I = 1/2ab log|ax-b/ax+b| + c

Question 3. Evaluate âˆ« 1/ a2x2+b2 dx

Solution:

Let us assume I = âˆ« 1/ a2x2+b2 dx

take 1/a2 common from above equation

= 1/a2 âˆ« 1/ x2+(b2/a2) dx

= 1/a2 âˆ« 1/ x2+(b/a)2 dx

Integrate the above eq then, we get

= (1/a2) 1/(b/a)tan-1[x/(b/a)] + c [since âˆ«1/ x2+a2 dx = 1/a tan-1(x/a) + c]

= 1/ab tan-1(ax/b) + c

Hence, I = 1/ab tan-1(ax/b) + c

Question 4. Evaluate âˆ« x2-1/ x2+4 dx

Solution:

Let us assume I = âˆ« x2-1/ x2+4 dx

We can write the above eq as below,

= âˆ« x2-1+4-4/ x2+4 dx

= âˆ« (x2+4)-4-1/ x2+4 dx

= âˆ« (x2+4)-5/ x2+4 dx

= âˆ« (x2+4)/ x2+4 dx – âˆ« 5/ x2+4 dx

= âˆ« dx – 5âˆ« 1/ x2+(2)2 dx

Integrate the above eq then, we get

= x – 5/2 tan-1(x/2) +c [since âˆ«1/ x2+a2 dx = 1/a tan-1(x/a) + c]

Hence I = x – 5/2 tan-1(x/2) +c

Question 5. Evaluate âˆ« 1/ âˆš1+4x2 dx

Solution:

Let us assume I = âˆ« 1/ âˆš1+4x2 dx

= âˆ« 1/ âˆš1+(2x)2 dx (i)

Let 2x = t

2dx = dt

Put the above value in eq (i)

=1/2 âˆ« 1/ âˆš1+t2 dt

Integrate the above eq then, we get

=1/2 log|t+âˆšt2+1| + c [since âˆ«1/ âˆšx2+a2 dx = log|x+âˆšx2+a2| +c]

=1/2 log|2x+âˆš(2x)2+1| + c

Hence, I =1/2 log|2x+âˆš4x2+1| + c

Question 6. Evaluate âˆ«1/ âˆša2+b2x2 dx

Solution:

Let us assume I = âˆ«1/ âˆša2+b2x2 dx

= âˆ«1/ âˆša2+(bx)2 dx (i)

Let bx = t

bdx = dt

dx = dt/b

Put the above value in eq (i)

= 1/b âˆ«1/ âˆša2+(t)2 dt

Integrate the above eq then, we get

= 1/b log|t+âˆša2+t2|+ c [since âˆ«1/ âˆša2+x2 dx = log|x+âˆša2+x2| + c]

= 1/b log|bx+âˆša2+(bx)2|+ c

Hence, I = 1/b log|bx+âˆša2+b2x2|+ c

Question 7. Evaluate âˆ«1/ âˆša2-b2x2 dx

Solution:

Let us assume I = âˆ«1/ âˆša2-b2x2 dx

= âˆ«1/ âˆša2-(bx)2 dx (i)

Let bx = t

bdx = dt

dx = dt/b

Put the above value in eq (i)

= 1/b âˆ«1/ âˆša2-(t)2 dt

Integrate the above eq then, we get

= 1/b sin-1(t/a)+ c [since âˆ«1/ âˆša2-x2 dx = sin-1 (x/a)+ c]

Hence, I = 1/b sin-1(bx/a)+ c

Question 8. Evaluate âˆ«1/ âˆš(2-x)2+1 dx

Solution:

Let us assume I = âˆ«1/ âˆš(2-x)2+1 dx (i)

Let 2-x=t

-dx = dt

Put the above value in eq (i)

= -âˆ«1/ âˆš(t)2+(1)2 dt

Integrate the above eq then, we get

= – log |t+âˆš(t)2+1| + c [since âˆ«1/ âˆšx2+a2 dx = log|x+âˆšx2+a2| +c]

= – log |(2-x)+âˆš(2-x)2+1| + c

Hence, I = – log |(2-x)+âˆš(2-x)2+1| + c

Question 9. Evaluate âˆ«1/ âˆš(2-x)2-1 dx

Solution:

Let us assume I = âˆ«1/ âˆš(2-x)2-1 dx (i)

Let 2-x=t

-dx = dt

Put the above value in eq (i)

= -âˆ«1/ âˆš(t)2-(1)2 dt

Integrate the above eq then, we get

= – log |t+âˆš(t)2-1| + c [since âˆ«1/ âˆšx2-a2 dx = log|x+âˆšx2-a2| +c]

= – log |(2-x)+âˆš(2-x)2-1| + c

Hence, I = – log |(2-x)+âˆš(2-x)2-1| + c

Question 10. Evaluate âˆ« x4+1/ x2+1 dx

Solution:

Let us assume I = âˆ« x4+1/ x2+1 dx

= âˆ« (x2)2+(1)2/ x2+1 dx

= âˆ« (x2+1)2-2x2/ x2+1 dx [a2 +b2 = (a+b)2-2ab]

= âˆ« (x2+1)2/ x2+1 dx -âˆ« 2x2/ x2+1 dx

= âˆ« (x2+1) dx – âˆ« (2x2+2-2/ x2+1) dx

= âˆ« (x2+1) dx – âˆ« 2(x2+1)/ x2+1 dx +âˆ« 2/ x2+1 dx

= âˆ« (x2+1) dx – âˆ« 2 dx + 2âˆ« 1/ x2+1 dx

Integrate the above eq then, we get

= x3/3 + x – 2x + 2tan-1(x) + c [since âˆ«1/ x2+a2 dx = 1/a tan-1(x/a) + c]

= x3/3 – x + 2tan-1(x) + c

Hence, I = x3/3 – x + 2tan-1(x) + c

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