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

### Question 1. Evaluate ∫ sec^{2}x/ 1 – tan^{2}x dx

**Solution:**

Let us assume I = ∫ sec

^{2}x/ 1 – tan^{2}x dx …..(i)Now, put tan x = t

sec

^{2}x dx = dtSo, put all these values in eq(i)

= ∫ dt/ 1

^{2 }– t^{2}On integrating the above equation then, we get

= 1/ 2(1) log|1 + t/1 – t| + c

Since, ∫ 1/ a

^{2 }– x^{2}dx = 1/ 2a log|a + x/a – x| + c]Hence, I = 1/2 log|1 + tanx/1 – tanx| + c

### Question 2. Evaluate ∫ e^{x}/ 1 + e^{2x} dx

**Solution:**

Let us assume I = ∫ e

^{x}/ 1 + e^{2x}dx …..(i)Now, put e

^{x}= te

^{x}dx = dtSo, put all these values in eq(i)

= dt/ 1 + t

^{2}On integrating the above equation then, we get

= tan

^{-1}t + cSince, ∫ 1/ 1 + x

^{2}dx = tan^{-1}x + cHence, I = tan

^{-1}e^{x}+ c

### Question 3. Evaluate ∫ cosx/ sin^{2}x + 4sinx + 5 dx

**Solution:**

Let us assume I = ∫ cosx/ sin

^{2}x + 4sinx + 5 dx …..(i)Now, put sinx = t

cosx dx = dt

So, put all these values in eq(i)

= ∫ dt/ t

^{2 }+ 4t + 5= ∫ dt/ t

^{2 }+ 2t(2) + (2)^{2 }– (2)^{2 }+ 5= ∫ dt/ (t + 2)

^{2 }+ 1 …..(ii)Again, Put t + 2 = u

dt = du

Now, put all these values in eq(ii)

= ∫ du/ u

^{2 }+ 1On integrating the above equation then, we get

= tan

^{-1}u + cSince, ∫1/ x

^{2 }+ 1 dx = tan^{-1}x + c= tan

^{-1}(t + 2) + cHence, I = tan

^{-1}(sinx + 2) + c

### Question 4. Evaluate ∫ e^{x}/e^{2x }+ 5e^{x }+ 6 dx

**Solution:**

Let us assume I = ∫ e

^{x}/e^{2x }+ 5e^{x }+ 6 dx …..(i)Now, put e

^{x}= te

^{x}dx = dtSo, put all these values in eq(i)

= ∫ dt/ t

^{2 }+ 5t + 6= ∫ dt/ t + 2t(5/2) + (5/2)

^{2 }– (5/2)^{2 }+ 6= ∫ dt/ (t + 5/2)

^{2 }– 1/4 …..(ii)Put t + 5/2 = u

dt = du

Now, put the above value in eq(ii)

= ∫ du/ u

^{2 }– (1/2)^{2}On integrating the above equation then, we get

= 2/2 log|u – (1/2)/u + (1/2)| + c

Since, ∫ 1/ x

^{2}– a^{2}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|e

^{x }+ 2/e^{x }+ 3| + c

### Question 5. Evaluate ∫ e^{3x}/ 4e^{6x }– 9 dx

**Solution:**

Let us assume I = ∫ e

^{3x}/ 4e^{6x}– 9 dx …..(i)Now, put e

^{3x}= t3e

^{3x}dx = dte

^{3x}dx = dt/3Now, put the above value in eq(i)

= 1/3 ∫ dt/ 4t

^{2 }– 9= 1/12 ∫ dt/ t

^{2 }– (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, ∫1/ x

^{2 }– a^{2}dx = 1/2a log|x – a/x + a| + c]= 1/36 log|2t – 3/2t + 3| + c

Hence, I = 1/36 log|2e

^{3x }– 3/2e^{3x }+ 3| + c

### Question 6. Evaluate ∫ dx/e^{x }+ e^{-x}

**Solution:**

Let us assume I = ∫ dx/e

^{x }+ e^{-x}=

^{ }∫^{ }dx/e^{x }+ 1/e^{x}= ∫ e

^{x}dx/ (e^{x})^{2}+ 1 …..(i)Now, put e

^{x}= te

^{x}dx = dtNow, put the above value in eq(i)

= ∫ dt/ t

^{2 }+ 1On integrating the above equation then, we get

= tan

^{-1}t + cSince ∫ 1/ 1 + x

^{2}dx = tan^{-1}x + cHence, I = tan

^{-1}(e^{x}) + c

### Question 7. Evaluate ∫ x/ x^{4 }+ 2x^{2 }+ 3 dx

**Solution:**

Let us assume I = ∫ x/ x

^{4 }+ 2x^{2 }+ 3 dx …..(i)Now, put x

^{2}= t2x dx = dt

x dx = dt/2

Now, put the above value in eq(i)

= 1/2 ∫ dt/ t

^{2 }+ 2t + 3= 1/2 ∫ dt/ t

^{2 }+ 2t + 1 – 1 + 3= 1/2 ∫ dt/ (t + 1)

^{2}+ 2 …..(ii)Now put t + 1 = u

dt = du

So, put the above value in eq(ii)

= 1/2 ∫ du/ u

^{2 }+ (√2)^{2}On integrating the above equation then, we get

= 1/2 x 1/√2 tan

^{-1}(u/√2) + cSince ∫1/ x

^{2 }+ a^{2}dx = 1/a tan^{-1}(x/a) + c= 1/2√2 tan

^{-1}(t + 1/ √2) + cHence, I = 1/2√2 tan

^{-1}(x^{2 }+ 1/ √2) + c

### Question 8. Evaluate ∫ 3x^{5}/ 1 + x^{12} dx

**Solution:**

Let us assume I = ∫ 3x

^{5}/ 1 + x^{12}dx= ∫ 3x

^{5}/ 1 + (x^{6})^{2}dx …..(i)Now, put x

^{6 }= t6x

^{5}dx = dtx

^{5}dx = dt/6Now, put the above value in eq(i)

= 3/6 ∫ dt/ 1 + t

^{2}On integrating the above equation then, we get

= 1/2 tan

^{-1}(t) + cSince ∫ 1/ x

^{2 }+ 1 dx = tan^{-1}x + cHence, I = 1/2 tan

^{-1}(x^{6}) + c

### Question 9. Evaluate ∫ x^{2}/ x^{6 }– a^{6} dx

**Solution:**

Let us assume I = ∫ x

^{2}/ x^{6 }– a^{6}dx= ∫ x

^{2}/ (x^{3})^{2 }– (a^{3})^{2}dx …..(i)Now, put x

^{3}= t3x

^{2}dx = dtx

^{2}dx = dt/3Now, put the above value in eq(i)

= 1/3 ∫ dt/ t

^{2 }– (a^{3})^{2}On integrating the above equation then, we get

= 1/3 x 1/2a

^{3}log|t – a^{3}/t + a^{3}| + cSince ∫1/ x

^{2 }– a^{2}dx = 1/2a log|x – a/x + a| + c= 1/6a

^{3}log|x^{3 }– a^{3}/x^{3 }+ a^{3}| + cHence, I = 1/6a

^{3}log|x^{3 }– a^{3}/x^{3 }+ a^{3}| + c

### Question 10. Evaluate ∫ x^{2}/ x^{6 }+ a^{6} dx

**Solution:**

Let us assume I = ∫ x

^{2}/ x^{6 }+ a^{6}dx= ∫ x

^{2}/ (x^{3})^{2 }+ (a^{3})^{2}dx …..(i)Now, put x

^{3}= t3x

^{2}dx = dtx

^{2}dx = dt/3Now, put the above value in eq(i)

= 1/3 ∫ dt/ t

^{2 }+ (a^{3})^{2}On integrating the above equation then, we get

= 1/3 x (1/a

^{3}) tan^{-1}(t/a^{3}) + cSince, ∫1/ x

^{2 }+ a^{2}dx = 1/a tan^{-1}(x/a) + cHence, I = 1/3a

^{3}tan^{-1}(x^{3}/a^{3}) + c

### Question 11. Evaluate ∫ 1/ x(x^{6 }+ 1) dx

**Solution:**

Let us assume I = ∫ 1/ x(x

^{6 }+ 1) dx= ∫ x

^{5}/ x^{6}(x^{6 }+ 1) dx …..(i)Now, put x

^{6}= t6x

^{5}dx = dtx

^{5}dx = dt/6Now, put the above value in eq(i)

= 1/6 ∫dt/ t(t + 1)

= 1/6 ∫dt/ t

^{2 }+ t= 1/6 ∫dt/ t

^{2 }+ 2t(1/2) + (1/2)^{2 }– (1/2)^{2}= 1/6 ∫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 ∫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 ∫ 1/ x

^{2 }– a^{2}dx = 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|x

^{6}/ x^{6 }+ 1| + c

### Question 12. Evaluate ∫ x/ (x^{4 }– x^{2 }+ 1) dx

**Solution:**

Let us assume I = ∫ x/ (x

^{4 }– x^{2 }+ 1) dx …..(i)Let x

^{2}= t2x dx = dt

x dx = dt/2

Now, put the above value in eq(i)

= 1/2 ∫dt/ t

^{2 }– t + 1= 1/2 ∫dt/ t

^{2 }– 2t(1/2) + (1/2)^{2 }– (1/2)^{2 }+ 1= 1/2 ∫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 ∫du/ (u)

^{2 }+ (√3/2)^{2}On integrating the above equation then, we get

= 1/2 x 1/(√3/2) tan

^{-1}(u/(√3/2)) + cSince, ∫ 1/ x

^{2 }+ a^{2}dx = 1/a tan^{-1}(x/a) + c= 1/√3 tan

^{-1}(t – 1/2/ (√3/2)) + cHence, I = 1/√3 tan

^{-1}(2x^{2 }– 1/ √3) + c

### Question 13. Evaluate ∫ x/ (3x^{4 }– 18x^{2 }+ 11) dx

**Solution:**

Let us assume I = ∫ x/ (3x

^{4 }– 18x^{2 }+ 11) dx= 1/3 ∫ x/ (x

^{4 }– 6x^{2 }+ 11/3) dx …..(i)Let x

^{2}= t2x dx = dt

x dx = dt/2

So, put the above value in eq(i)

= 1/3 x 1/2 ∫dt/ t

^{2 }– 6t + 11/3= 1/6 ∫dt/ t

^{2 }– 2t(3) + (3)^{2 }– (3)^{2 }+ 11/3= 1/6 ∫dt/ (t – 3)

^{2 }– (16/3) …..(ii)Let t – 3 = u

dt = du

Now, put the above value in eq(ii)

= 1/6 ∫du/ (u)

^{2 }– (4/√3)^{2}On integrating the above equation then, we get

= 1/6 x 1/ 2(4/√3) log|u – (4/√3)/u + (4/√3)| + c

Since, ∫ 1/ x

^{2 }– a^{2}dx = 1/2a log|x – a/x + a| + c= √3/48 log|(t – 3 – 4/√3)/(t – 3 + 4/√3)| + c

Hence, I = √3/48 log|(x

^{2 }– 3 – 4/√3)/(x^{2 }– 3 + 4/√3)| + c

### Question 14. Evaluate ∫ e^{x}/ (1 + e^{x})(2 + e^{x}) dx

**Solution:**

Let us assume I = ∫ e

^{x}/ (1 + e^{x})(2 + e^{x}) dx …..(i)Let e

^{x}= te

^{x}dx = dtSo, put the above value in eq(i)

= ∫ dt/ (1 + t)(2 + t)

= ∫dt/ (1 + t) – ∫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 + e

^{x}/2 + e^{x}| + c

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