Get Answers to all your Questions

header-bg qa

Explain solution rd sharma class 12 chapter Indefinite Integrals exercise 18.2 question 39

Answers (1)

\begin{aligned} &\frac{x^{3}}{3}+\frac{x^{2}}{2}-2 x+c\\ &\text { Hint: Using } \int x^{n} d x\\ &\text { Given: } \int \frac{\left(x^{3}+8\right)(x-1)}{x^{2}-2 x+4} d x \end{aligned}\begin{aligned} &\text { Solution: } I=\int \frac{\left(x^{3}+8\right)(x-1)}{x^{2}-2 x+4} d x \\ &=\int \frac{\left(x^{3}+2^{3}\right)(x-1)}{x^{2}-2 x+4} d x \\ &=\int \frac{(x+2)\left(x^{2}-2 x+4\right)(x-1)}{x^{2}-2 x+4} d x \quad\left[\because a^{3}+b^{3}=(a+b)\left(a^{2}+b^{2}-a b\right)\right] \\ &=\int(x+2)(x-1) d x \\ &=\int\left(x^{2}+x-2\right) d x \quad\left[\begin{array}{l} \because(x+2)(x-1)=x(x-1)+2(x-1) \\ =x^{2}-x+2 x-2=x^{2}+x-2 \end{array}\right] \\ &=\int x^{2} d x+\int x d x+2 \int 1 d x \\ &=\frac{x^{3}}{3}+\frac{x^{2}}{2}-2 x+c \end{aligned}

Posted by

Info Expert 29

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support