Class 12 RD Sharma Solutions Chapter 19 Indefinite Integrals Exercise 19.4 (original) (raw)

Last Updated : 4 Sep, 2024

Question 1. Integrate ∫ \frac{(x^2 + 5x +2)}{(x + 2)} dx

**Solution:

Let, I = ∫ \frac{(x^2 + 5x +2)}{(x + 2)} dx

Use division method, then we get,

\frac{(x2 + 5x +2)}{(x+2)} = x + 3 - \frac{4}{(x+2)}

∫ \frac{(x^2 + 5x +2)}{(x + 2)} =∫ x + 3 - \frac{4}{(x+2)} dx

= ∫ (x + 3)dx - 4∫1/(x + 2) dx

Integrate the above equation, then we get

= x2/2 + 3x - 4 log |x + 2| + c

Hence, I = x2/2 + 3x - 4 log |x + 2| + c

Question 2. Integrate ∫ \frac{x^3}{(x-2)} dx

**Solution:

Let I = ∫ \frac{x^3}{(x-2)} dx

Use division method, then we get,

\frac{x^3}{(x+2)} =x^2 + 2x + 4 + \frac{8}{(x-2)}

= ∫ x2 dx + 2∫x dx + 4∫ dx + 8 ∫1/(x - 2) dx

Integrate the above equation, then we get

= x3/3 + 2x2/2 + 4x + 8 log|x - 2| + c

= x3/3 + x2 + 4x + 8 log|x - 2| + c

Hence, I = x3/3 + x2 + 4x + 8 log|x - 2| + c

Question 3. Integrate ∫ \frac{(x^2 + x + 5)}{(3x +2)} dx

**Solution:

Let, I = ∫ \frac{(x^2 + x + 5)}{(3x +2)} dx

Use division method, then we get,

\frac{(x^2 + x + 5)}{(3x +2)} = \frac{x}{3} + \frac{1}{9} + \frac{43}{9}\times \frac{1}{3x+2} dx

= ∫\frac{x}{3}dx + \frac{1}{9}∫1dx + \frac{43}{9} ∫\frac{1}{(3x+2)} dx

Integrate the above equation, then we get

= \frac{x^2}{6} + \frac{x}{9} + \frac{43}{9} \frac{1}{3} log|3x+2| + c

= x2 /6 + x/9 + (43/27) log|3x + 2| + c

Hence, I = = x2 /6 + x/9 + (43/27) log|3x + 2| + c

Question 4. Integrate ∫ \frac{(2x+3)}{(x-1)^2} dx

**Solution:

Let, I = ∫ \frac{(2x+3)}{(x-1)^2} dx

We can write the above equation as below,

= ∫ \frac{(2x + 2 - 2 + 3)}{(x-1)^2} dx

On solving the above equation,

= ∫ \frac{(2x - 2 + 5)}{(x-1)^2} dx

= ∫ \frac{2(x - 1)}{(x-1)^2} dx + 5 ∫\frac{1}{(x-1)^2} dx

= 2∫ (1/(x - 1) dx + 5 ∫(x - 1)-2 dx

Integrate the above equation, then we get

= 2 log|x - 1| + 5 (x - 1)-1/(-1) + c

= 2 log|x - 1| - 5 / (x - 1) + c

Hence, I = 2 log|x - 1| - 5 / (x - 1) + c

Question 5. Integrate ∫ \frac{(x^2 + 3x - 1)}{(x+1)^2} dx

**Solution:

Let, I = ∫ \frac{(x^2 + 3x - 1)}{(x+1)^2} dx

We can write the above equation as below,

= ∫\frac{(x^2 + x + 2x - 1)}{(x+1)^2} dx

= ∫\frac{x(x+1)}{(x+1)^2} dx + ∫\frac{(2x - 1)}{(x+1)^2} dx

= ∫\frac{x}{(x+1)} dx + ∫\frac{\sqrt{2x + 2 - 2 + 1}}{(x+1)^2} dx

= ∫\frac{(x+1)}{(x+1)} dx - ∫\frac{1}{(x+1)} dx + 2∫\frac{(x + 1)}{(x+1)^2} dx - 3∫\frac{1}{(x+1)^2} dx

= ∫ dx - ∫ 1/ (x + 1) dx + 2∫1/(x + 1) dx -3 ∫(x + 1)-2 dx

Integrate the above equation, then we get

= x - log|x + 1| + 2 log|x + 1| - 3(x + 1)-1/(-1) + c

= x - log|x + 1| + 2 log|x + 1| + 3/(x + 1) + c

= x + log|x + 1| + 3/(x + 1) + c

Hence, I = x + log|x + 1| + 3/(x + 1) + c

Question 6. Integrate ∫ \frac{(2x - 1)}{(x - 1)^2} dx

**Solution:

Let, I =∫ \frac{(2x - 1)}{(x - 1)^2} dx

= ∫ \frac{(2x - 1 + 2 - 2)}{(x - 1)^2} dx

= ∫ \frac{(2x - 2 + 1)}{(x - 1)^2} dx

= ∫ \frac{(2x - 2)}{(x - 1)^2} dx+ ∫\frac{1}{(x - 1)^2} dx

= 2∫ \frac{(x - 1)}{(x - 1)^2} dx+ ∫\frac{1}{(x - 1)^2} dx

= 2∫ 1/(x - 1) dx + ∫(x - 1)-2 dx

Integrate the above equation, then we get

= 2 log|x - 1| + (x - 1)-1/(-1) + c

= 2 log|x - 1| - 1/(x - 1) + c

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

Summary

Exercise 19.4 in Chapter 19 on Indefinite Integrals focuses on integrating functions involving algebraic expressions raised to various powers. This exercise primarily employs the substitution method for integration. The practice questions cover a range of scenarios, including positive and negative integer exponents, as well as fractional exponents. Students are required to recognize the appropriate substitution, perform the integration, and then revert back to the original variable. The problems progress in complexity, starting with simpler polynomial expressions and moving to more complex algebraic expressions involving square roots and higher powers. This exercise aims to strengthen students' ability to apply the substitution method effectively and enhance their overall integration skills.