Exams Joint Entrance Examination Mathematics Indefinite Integrals Question 291 Question 291 of 416 Medium MCQ
The integral ∫ 2 x 12 + 5 x 9 ( x 5 + x 3 + 1 ) 3 d x \displaystyle\int \dfrac{2x^{12}+5x^9}{(x^5+x^3+1)^3}\,dx ∫ ( x 5 + x 3 + 1 ) 3 2 x 12 + 5 x 9 d x is equal to:
Solutions · 1 Post your solution Official solution
Dividing numerator and denominator by x 15 x^{15} x 15 : ∫ ( 2 x 3 + 5 x 6 ) d x ( 1 + 1 x 2 + 1 x 5 ) 3 \int \dfrac{\left(\dfrac{2}{x^3}+\dfrac{5}{x^6}\right)dx}{\left(1+\dfrac{1}{x^2}+\dfrac{1}{x^5}\right)^3} ∫ ( 1 + x 2 1 + x 5 1 ) 3 ( x 3 2 + x 6 5 ) d x
Let 1 + 1 x 2 + 1 x 5 = t ⇒ ( − 2 x 3 − 5 x 6 ) d x = d t 1+\dfrac{1}{x^2}+\dfrac{1}{x^5}=t \Rightarrow \left(-\dfrac{2}{x^3}-\dfrac{5}{x^6}\right)dx = dt 1 + x 2 1 + x 5 1 = t ⇒ ( − x 3 2 − x 6 5 ) d x = d t
∫ − d t t 3 = 1 2 t 2 + C = 1 2 ( 1 + 1 x 2 + 1 x 5 ) 2 + C = x 10 2 ( x 5 + x 3 + 1 ) 2 + C \int \dfrac{-dt}{t^3} = \dfrac{1}{2t^2}+C = \dfrac{1}{2\left(1+\dfrac{1}{x^2}+\dfrac{1}{x^5}\right)^2}+C = \dfrac{x^{10}}{2(x^5+x^3+1)^2}+C ∫ t 3 − d t = 2 t 2 1 + C = 2 ( 1 + x 2 1 + x 5 1 ) 2 1 + C = 2 ( x 5 + x 3 + 1 ) 2 x 10 + C Answer the question to unlock