# Class 12 RD Sharma Solutions – Chapter 19 Indefinite Integrals – Exercise 19.16

Last Updated : 19 Apr, 2021

### Question 1. Evaluate âˆ« sec2x/ 1 – tan2x dx

Solution:

Let us assume I = âˆ« sec2x/ 1 – tan2x dx              …..(i)

Now, put tan x = t

sec2x dx = dt

So, put all these values in eq(i)

= âˆ« dt/ 12 – t2

On integrating the above equation then, we get

= 1/ 2(1) log|1 + t/1 – t| + c

Since, âˆ« 1/ a2 – x2 dx = 1/ 2a log|a + x/a – x| + c]

Hence, I = 1/2 log|1 + tanx/1 – tanx| + c

### Question 2. Evaluate âˆ« ex/ 1 + e2x dx

Solution:

Let us assume I = âˆ« ex/ 1 + e2x dx              …..(i)

Now, put ex = t

ex dx = dt

So, put all these values in eq(i)

= dt/ 1 + t2

On integrating the above equation then, we get

= tan-1t + c

Since, âˆ« 1/ 1 + x2dx = tan-1x + c

Hence, I = tan-1ex + c

### Question 3. Evaluate âˆ« cosx/ sin2x + 4sinx + 5 dx

Solution:

Let us assume I = âˆ« cosx/ sin2x + 4sinx + 5 dx               …..(i)

Now, put sinx = t

cosx dx = dt

So, put all these values in eq(i)

= âˆ« dt/ t2 + 4t + 5

= âˆ« dt/ t2 + 2t(2) + (2)2 – (2)2 + 5

= âˆ« dt/ (t + 2)2 + 1                      …..(ii)

Again, Put t + 2 = u

dt = du

Now, put all these values in eq(ii)

= âˆ« du/ u2 + 1

On integrating the above equation then, we get

= tan-1u + c

Since, âˆ«1/ x2 + 1 dx = tan-1x + c

= tan-1(t + 2) + c

Hence, I = tan-1(sinx + 2) + c

### Question 4. Evaluate âˆ« ex/e2x + 5ex + 6 dx

Solution:

Let us assume I = âˆ« ex/e2x + 5ex + 6 dx                  …..(i)

Now, put ex = t

ex dx = dt

So, put all these values in eq(i)

= âˆ« dt/ t2 + 5t + 6

= âˆ« dt/ t + 2t(5/2) + (5/2)2 – (5/2)2 + 6

= âˆ« dt/ (t + 5/2)2 – 1/4                      …..(ii)

Put t + 5/2 = u

dt = du

Now, put the above value in eq(ii)

= âˆ« du/ u2 – (1/2)2

On integrating the above equation then, we get

= 2/2 log|u – (1/2)/u + (1/2)| + c

Since, âˆ« 1/ x2 – a2 dx = 1/ 2alog|x – a/x + a| + c

= log|2u – 1/2u + 1| + c

= log|2(t + 5/2) – 1/2(t + 5/2) + 1| + c

Hence, I = log|ex + 2/ex + 3| + c

### Question 5. Evaluate âˆ« e3x/ 4e6x – 9 dx

Solution:

Let us assume I = âˆ« e3x/ 4e6x– 9 dx                …..(i)

Now, put e3x = t

3e3x dx = dt

e3x dx = dt/3

Now, put the above value in eq(i)

= 1/3 âˆ« dt/ 4t2 – 9

= 1/12 âˆ« dt/ t2 – (3/2)2

On integrating the above equation then, we get

= 1/12 x 1/ 2(3/2) log|t – 3/2/t + 3/2| + c

Since, âˆ«1/ x2 – a2 dx = 1/2a log|x – a/x + a| + c]

= 1/36 log|2t – 3/2t + 3| + c

Hence, I = 1/36 log|2e3x – 3/2e3x + 3| + c

### Question 6. Evaluate âˆ« dx/ex + e-x

Solution:

Let us assume I = âˆ« dx/ex + e-x

= âˆ« dx/ex + 1/ex

= âˆ« exdx/ (ex)2 + 1                   …..(i)

Now, put ex = t

exdx = dt

Now, put the above value in eq(i)

= âˆ« dt/ t2 + 1

On integrating the above equation then, we get

= tan-1t + c

Since âˆ« 1/ 1 + x2 dx = tan-1x + c

Hence, I = tan-1(ex) + c

### Question 7. Evaluate âˆ« x/ x4 + 2x2 + 3 dx

Solution:

Let us assume I = âˆ« x/ x4 + 2x2 + 3 dx               …..(i)

Now, put x2 = t

2x dx = dt

x dx = dt/2

Now, put the above value in eq(i)

= 1/2 âˆ« dt/ t2 + 2t + 3

= 1/2 âˆ« dt/ t2 + 2t + 1 – 1 + 3

= 1/2 âˆ« dt/ (t + 1)2 + 2                         …..(ii)

Now put t + 1 = u

dt = du

So, put the above value in eq(ii)

= 1/2 âˆ« du/ u2 + (âˆš2)2

On integrating the above equation then, we get

= 1/2 x 1/âˆš2 tan-1(u/âˆš2) + c

Since âˆ«1/ x2 + a2dx = 1/a tan-1(x/a) + c

= 1/2âˆš2 tan-1(t + 1/ âˆš2) + c

Hence, I = 1/2âˆš2 tan-1(x2 + 1/ âˆš2) + c

### Question 8. Evaluate âˆ« 3x5/ 1 + x12 dx

Solution:

Let us assume I = âˆ« 3x5/ 1 + x12 dx

= âˆ« 3x5/ 1 + (x6)2dx                 …..(i)

Now, put x6 = t

6x5dx = dt

x5dx = dt/6

Now, put the above value in eq(i)

= 3/6 âˆ« dt/ 1 + t2

On integrating the above equation then, we get

= 1/2 tan-1(t) + c

Since âˆ« 1/ x2 + 1 dx = tan-1x + c

Hence, I = 1/2 tan-1(x6) + c

### Question 9. Evaluate âˆ« x2/ x6 – a6 dx

Solution:

Let us assume I = âˆ« x2/ x6 – a6 dx

= âˆ« x2/ (x3)2 – (a3)2 dx                …..(i)

Now, put x3 = t

3x2 dx = dt

x2 dx = dt/3

Now, put the above value in eq(i)

= 1/3 âˆ« dt/ t2 – (a3)2

On integrating the above equation then, we get

= 1/3 x 1/2a3 log|t – a3/t + a3| + c

Since âˆ«1/ x2 – a2 dx = 1/2a log|x – a/x + a| + c

= 1/6a3 log|x3 – a3/x3 + a3| + c

Hence, I = 1/6a3 log|x3 – a3/x3 + a3| + c

### Question 10. Evaluate âˆ« x2/ x6 + a6 dx

Solution:

Let us assume I = âˆ« x2/ x6 + a6 dx

= âˆ« x2/ (x3)2 + (a3)2 dx                 …..(i)

Now, put x3 = t

3x2 dx = dt

x2 dx = dt/3

Now, put the above value in eq(i)

= 1/3 âˆ« dt/ t2 + (a3)2

On integrating the above equation then, we get

= 1/3 x (1/a3) tan-1(t/a3) + c

Since, âˆ«1/ x2 + a2 dx = 1/a tan-1(x/a) + c

Hence, I = 1/3a3 tan-1(x3/a3) + c

### Question 11. Evaluate âˆ« 1/ x(x6 + 1) dx

Solution:

Let us assume I = âˆ« 1/ x(x6 + 1) dx

= âˆ« x5/ x6(x6 + 1) dx                     …..(i)

Now, put x6 = t

6x5 dx = dt

x5 dx = dt/6

Now, put the above value in eq(i)

= 1/6 âˆ«dt/ t(t + 1)

= 1/6 âˆ«dt/ t2 + t

= 1/6 âˆ«dt/ t2 + 2t(1/2) + (1/2)2 – (1/2)2

= 1/6 âˆ«dt/ (t + 1/2)2 – (1/2)2                 …..(ii)

Let t + 1/2 = u

dt = du

So, put the above value in eq(ii)

= 1/6 âˆ«du/ (u)2 – (1/2)2

On integrating the above equation then, we get

= 1/6 x 1/ 2(1/2) log|u – (1/2)/u + (1/2)| + c

Since âˆ« 1/ x2 – a2dx = 1/2a log|x – a/x + a| + c

= 1/6 log|{(t + 1/2) – 1/2}/(t + 1/2) + 1/2| + c

Hence, I = 1/6 log|x6/ x6 + 1| + c

### Question 12. Evaluate âˆ« x/ (x4 – x2 + 1) dx

Solution:

Let us assume I = âˆ« x/ (x4 – x2 + 1) dx                  …..(i)

Let x2 = t

2x dx = dt

x dx = dt/2

Now, put the above value in eq(i)

= 1/2 âˆ«dt/ t2 – t + 1

= 1/2 âˆ«dt/ t2 – 2t(1/2) + (1/2)2 – (1/2)2 + 1

= 1/2 âˆ«dt/ (t – 1/2)2 + (3/4)                   …..(ii)

Let t – 1/2 = u

dt = du

So, put the above value in eq(ii)

= 1/2 âˆ«du/ (u)2 + (âˆš3/2)2

On integrating the above equation then, we get

= 1/2 x 1/(âˆš3/2) tan-1(u/(âˆš3/2)) + c

Since, âˆ« 1/ x2 + a2dx = 1/a tan-1(x/a) + c

= 1/âˆš3 tan-1(t – 1/2/ (âˆš3/2)) + c

Hence, I = 1/âˆš3 tan-1(2x2 – 1/ âˆš3) + c

### Question 13. Evaluate âˆ« x/ (3x4 – 18x2 + 11) dx

Solution:

Let us assume I = âˆ« x/ (3x4 – 18x2 + 11) dx

= 1/3 âˆ« x/ (x4 – 6x2 + 11/3) dx                   …..(i)

Let x2 = t

2x dx = dt

x dx = dt/2

So, put the above value in eq(i)

= 1/3 x 1/2 âˆ«dt/ t2 – 6t + 11/3

= 1/6 âˆ«dt/ t2 – 2t(3) + (3)2 – (3)2 + 11/3

= 1/6 âˆ«dt/ (t – 3)2 – (16/3)                      …..(ii)

Let t – 3 = u

dt = du

Now, put the above value in eq(ii)

= 1/6 âˆ«du/ (u)2 – (4/âˆš3)2

On integrating the above equation then, we get

= 1/6 x 1/ 2(4/âˆš3) log|u – (4/âˆš3)/u + (4/âˆš3)| + c

Since, âˆ« 1/ x2 – a2dx = 1/2a log|x – a/x + a| + c

= âˆš3/48 log|(t – 3 – 4/âˆš3)/(t – 3 + 4/âˆš3)| + c

Hence, I = âˆš3/48 log|(x2 – 3 – 4/âˆš3)/(x2 – 3 + 4/âˆš3)| + c

### Question 14. Evaluate âˆ« ex/ (1 + ex)(2 + ex) dx

Solution:

Let us assume I = âˆ« ex/ (1 + ex)(2 + ex) dx                       …..(i)

Let ex = t

ex dx = dt

So, put the above value in eq(i)

= âˆ« dt/ (1 + t)(2 + t)

= âˆ«dt/ (1 + t) – âˆ«dt/(2 + t)

On integrating the above equation then, we get

= log|1 + t| – log|2 + t| + c

= log|1 + t/2 + t| + c

Hence, I = log|1 + ex/2 + ex| + c

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