# Class 12 RD Sharma Solutions – Chapter 19 Indefinite Integrals – Exercise 19.2 | Set 1

### Question 1. Evaluate (∫(3x√5 + 4√x + 5)dx

**Solution:**

We have, (∫(3x√5+4√x+5)dx

= ∫3x√5 dx + ∫4√x dx + ∫5dx

= ∫3x

^{3/2}dx + 4∫x^{1/2}dx + 5∫dx= x

^{(3/2)+1}/(3/2 + 1) + 4x^{(1/2)+1}/(1/2 + 1) + 5x + c= 6/5 x

^{5/2 }+ 8/3 x^{3/2 }+ 5x + c

### Question 2. Evaluate ∫(2^{x }+ 5/x – 1/x^{1/3})dx

**Solution:**

We have, ∫(2

^{x }+ 5/x – 1/x^{1/3})dx= ∫2

^{x}dx + 5∫1/x dx – ∫1/x^{1/3 }dx= 2

^{x}/(log2) + 5logx – 3/2 x^{2/3 }+ c

### Question 3. Evaluate ∫{√x (ax^{2 }+ bx + c)}dx

**Solution:**

We have, ∫{√x (ax

^{2 }+ bx + c)}dx= ∫√x × ax

^{2}dx + ∫√x × bx dx + ∫c√x dx= ∫ax

^{5/2}dx + ∫bx^{3/2}dx + ∫cx^{1/2}dx= (ax

^{(5/2)+1})/(5/2 + 1) + (bx^{(3/2)+1})/(3/2 + 1) + (cx^{(1/2)+1})/(1/2 + 1) + d= (2ax

^{7/2})/7 + (2bx^{5/2})/5 + (2cx^{3/2})/3 + d

### Question 4. Evaluate ∫(2 – 3x) (3 + 2x)(1 – 2x)dx

**Solution:**

We have, ∫(2 – 3x) (3 + 2x)(1 – 2x)dx

= ∫(6 + 4x – 9x – 6x

^{2})(1 – 2x)dx= ∫(-6x

^{2 }– 5x + 6)(1 – 2x)dx= ∫(-6x

^{2 }+ 12x^{3 }– 5x + 10x^{2 }+ 6 – 12x)dx= ∫(4x

^{2 }+ 12x^{3 }– 17x + 6)dx= ∫(12x

^{3 }+ 4x^{2 }– 17x + 6)dx= 12/4 x

^{4 }+ 4/3 x^{3 }– 17/2 x^{2 }+ 6x + c= 3x

^{4 }+ 4/3 x^{3 }– 17/2 x^{2 }+ 6x + c

### Question 5. Evaluate ∫(m/x + x/m + m^{x }+ x^{m }+ mx)dx

**Solution:**

We have, ∫(m/x + x/m + m

^{x }+ x^{m }+ mx)dx= m∫1/x dx + 1/m ∫xdx + ∫m

^{x}dx + ⌋x^{m}dx + m∫xdx= mlog|x| + x

^{2}/2m + m^{x}/(logm) + x^{m+1}/(m + 1) + (mx^{2})/2 + c

### Question 6. Evaluate ∫ (√x – 1/√x)^{2} dx

**Solution:**

We have, ∫ (√x – 1/√x)

^{2}dxBy using formula (x + y)

^{2}= x^{2}+ y^{2}+2xyWe get, ∫(x + 1/x – 2)dx

= ∫xdx + ∫1/x dx – 2∫1.dx

= x

^{2}/2 + log|x| – 2x + C

### Question 7. Evaluate ∫((1 + x)^{3})/√xdx

**Solution:**

We have, ∫((1 + x)

^{3})/√xdxBy using formula (x + y)

^{3}= x^{3}+ y^{3}+3x^{2}y + 3xy^{2}We get, ∫(1 + x

^{3 }+ 3x^{2 }+ 3x)/√x dx= 1/√x dx + ∫x

^{3}/√x dx + ∫(3x^{2})/√x dx + ∫3x/√x dx= ∫x

^{-1/2}dx + ∫x^{5/2}dx + 3∫x^{3/2}dx + 3∫x^{1/2}dx= x

^{(-1/2)+1}/((-1)/2 + 1) + (x^{(5/2)+1})/(5/2 + 1) + (3x^{(3/2)+1})/(3/2 + 1) + 3 x^{(1/2)+1}/(1/2 + 1) + c= x

^{1/2}/(1/2) + x^{7/2}/(7/2) + (3x^{5/2})/(5/2) + 3 x^{3/2}/(3/2) + c= 2x

^{1/2 }+ 2/7 x^{7/2 }+ 6/5 x^{5/2 }+ 6/3 x^{3/2 }+ c= 2x

^{1/2 }+ 2/7 x^{7/2 }+ 6/5 x^{5/2 }+ 2x^{3/2 }+ c

### Question 8. Evaluate ∫{x^{2 }+ e^{logx }+ (e/2)^{x} }dx

**Solution:**

We have, ∫{x

^{2 }+ e^{logx }+ (e/2)^{x}}dx= ∫x

^{2}dx + ∫e^{logx}dx + ∫(e/2)^{x}dx= x

^{3}/3 + ∫xdx + ∫(e/2)^{x}dx= x

^{3}/3 + x^{2}/2 + 1/(log(e/2)) × (e/2)^{x }+ c

### Question 9. Evaluate ∫ (x^{e }+ e^{x }+ e^{e})dx

**Solution:**

We have, ∫ (x

^{e }+ e^{x }+ e^{e})dx= ∫x

^{e}dx + ∫e^{x}dx + ∫e^{e}dx= x

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

### Question 10. Evaluate ∫√x (x^{3 }– 2/x)dx

**Solution:**

We have, ∫√x (x

^{3 }– 2/x)dx= ∫ x

^{7/2}dx – 2∫ x^{-1/2}dx= x

^{(7/2)+1}/(7/2 + 1) – 2 x^{(-1/2)+1}/((-1)/2 + 1) + c= x

^{9/2}/(9/2) – (2x^{-1/2})/((-1)/2) + c= 2/9 x

^{9/2 }– 4x^{-1/2 }+ c

### Question 11. Evaluate ∫1/√x (1 + 1/x)dx

**Solution:**

We have, ∫1/√x (1 + 1/x)dx

= ∫ (1/√x + 1/(√x × x))dx

= ∫x

^{-1/2 }+ ∫x^{-3/2}dx= 2x

^{1/2 }– 2x^{-1/2 }+ c

### Question 12. Evaluate ∫(x^{6 }+ 1)/(x^{2 }+ 1) dx

**Solution:**

We have, ∫(x

^{6 }+ 1)/(x^{2 }+ 1) dx= ∫((x

^{2})^{3 }+ (1)^{3})/(x^{2 }+ 1) dx= ∫(x

^{2 }+ 1)(x^{4 }+ 1 – x^{2})/(x^{2 }+ 1) dx= ∫(x

^{4 }– x^{2 }+ 1)dx= ∫x

^{4}dx – ∫x^{2}dx + ∫1dx= x

^{5}/5 – x^{3}/3 + x + c

### Question 13. Evaluate ∫ (x^{-1/3 }+ √x + 2)/∛x dx

**Solution:**

We have, (x

^{-1/3 }+ √x + 2)/∛x dx= ∫(x

^{-1/3}dx)/x^{1/3}+ ∫x^{1/2}/x^{1/3}dx + ∫2/x^{1/3}dx= ∫x

^{-2/3}dx + ∫x^{1/6}dx + 2∫ x^{-1/3}dx= 3x

^{1/3 }+ 6/7 x^{7/6 }+ 3x^{2/3 }+ c

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

**Solution:**

We have, ∫((1 + √x)

^{2})/√x dx= ∫(1 + x + 2√x)/x

^{1/2}dx= ∫x

^{-1/2}+∫ x^{1/2}dx + 2∫dx= 2x

^{1/2 }+ 2/3 x^{3/2 }+ 2x + c= 2√x + 2x + 2/3 x

^{3/2 }+ c

### Question 15. Evaluate ∫√x(3 – 5x)dx

**Solution:**

We have, ∫√x(3 – 5x)dx

= 3∫√x dx – 5x

^{3/2}dx= 3x

^{3/2}/(3/2) – 5 x^{5/2}/(5/2) + c= 2x

^{3/2 }– 2x^{5/2 }+ c

### Question 16. Evaluate ∫((x + 1)(x – 2))/√x dx

**Solution:**

We have, ∫((x + 1)(x – 2))/√x dx

= ∫(x

^{2 }– 2x + x – 2)/x^{1/2}dx= ∫(x

^{2 }– x – 2)/x^{1/2}dx= ∫x

^{2}/x^{1/2}dx – ∫x^{1/2}dx – 2∫x^{-1/2}dx= (2x

^{5/2})/5 – (2x^{3/2})/3 – 4x^{1/2 }+ c= 2/5 x

^{5/2 }– (2x^{3/2})/3 – 4√x + c

### Question 17. Evaluate ∫(x^{5 }+ x^{-2 }+ 2)/x^{2}dx

**Solution:**

We have, ∫(x

^{5 }+ x^{-2 }+ 2)/x^{2}dx= ∫(x

^{5}/x^{2}+x^{-2}/x^{2}+2/x^{2})dx= ∫x

^{3}dx + ∫x^{-4 }+ 2∫x^{-2}dx= x

^{4}/4 + x^{-3}/(-3) + (2x^{-1})/(-1) + c= x

^{4}/4 – x^{-3}/3 – 2/x + c

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

**Solution:**

We have, ∫(3x + 4)

^{2}dxBy using formula (x + y)

^{2}= x^{2}+ y^{2}+2xyWe get, ∫ (9x

^{2 }+ 16 + 24x)dx= ∫9x

^{2}dx + ∫16dx + ∫24xdx= 9 x

^{3}/3 + 16x + 24 x^{2}/2 + c= 3x

^{3 }+ 16x + 12x^{2 }+ c

### Question 19. Evaluate ∫(2x^{4 }+ 7x^{3 }+ 6x^{2})/(x^{2 }+ 2x) dx

**Solution:**

We have, ∫(2x

^{4 }+ 7x^{3 }+ 6x^{2})/(x^{2 }+ 2x) dx= ∫x(2x

^{3 }+ 7x^{2 }+ 6x)/(x(x + 2))dx= ∫(2x

^{3 }+ 7x^{2 }+ 6x)/(x + 2)dx= ∫ (2x

^{3 }+ 4x^{2 }+ 3x^{2 }+ 6x)/((x + 2))dx= ∫(2x

^{2}(x + 2) + 3x(x + 2))/(x + 2) dx= ∫(x + 2)(2x

^{2 }+ 3x)/(x + 2) dx= ∫(2x

^{2 }+ 3x)dx= ∫2x

^{2}dx + ∫3xdx= 2/3 x

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

### Question 20. Evaluate ∫(5x^{4 }+ 12x^{3 }+ 7x^{2})/(x^{2 }+ x) dx

**Solution:**

We have, ∫(5x

^{4 }+ 7x^{3 }+ 5x^{3 }+ 7x^{2})/(x^{2 }+ x) dx= ∫(5x

^{3 }+ 7x^{2 }+ 5x^{2 }+ 7x)/(x + 1) dx= ∫5x

^{2}(x + 1) + 7x(x + 1)/(x + 1) dx= ∫(5x

^{2 }+ 7x)dx= (5x

^{3})/3 + (7x^{2})/2 + C

### Question 21. Evaluate ∫(sin^{2}x)/(1 + cosx) dx

**Solution:**

We have, ∫(sin

^{2}x)/(1 + cosx) dx= ∫(1 – cos

^{2}x)/(1 + cosx) dx= ∫((1 – cosx)(1 + cosx))/(1 + cosx) dx

= ∫(1 – cosx)dx

= x – sinx + c

### Question 22. Evaluate ∫(sec^{2}x + cosec^{2} x)dx

**Solution:**

We have, ∫(sec

^{2}x + cosec^{2}x)dx= ∫ sec

^{2}xdx + ∫ cosec^{2}xdx= tanx – cotx + c

= tanx – cotx + c

### Question 23. Evaluate ∫(sin^{3}x – cos^{3}x)/(sin^{2}xcos^{2}x) dx

**Solution:**

We have, ∫(sin

^{3}x – cos^{3}x)/(sin^{2}xcos^{2}x) dx= ∫((sin

^{3}x)/(sin^{2}x cos^{2}x) – (cos^{3}x)/(sin^{2}x cos^{2}x))dx= ∫ (sinxsec

^{2}x – cosxcosec^{2}x)dx= ∫ (tanxsecx – cotxcosecx)dx

= secx + cosecx + c

### Question 24. Evaluate ∫(5cos^{3}x + 6sin^{3}x)/(2sin^{2}xcos^{2}x) dx

**Solution:**

We have, ∫(5cos

^{3}x + 6sin^{3}x)/(2sin^{2}xcos^{2}x) dx= ∫(5cos

^{3}x)/(2sin^{2}xcos^{2}x) dx + ∫(6sin^{3}x)/(2sin^{2}xcos^{2}x) dx= 5/2 ∫(cosx)/(sin

^{2}x) dx + 3∫(sinx)/(cos^{2}x) dx= 5/2 ∫cotxcosecxdx + 3∫ secxtanxdx

= (-5)/2 cosecx + 3secx+c

= (-5)/2 cossecx + 3secx+c

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