Question

Which of these integrals is equivalent to the given integral?

∫ x^3/(√9x^4+6x^2-1) dx

a. ∫ x^3/(3x^2+1) dx
b. 1/6 ∫ √ secθ-1 /√2tanθ dθ

Answer 1

`\mathbf{\int (x^3)/(√(9x^4+ 6x^2 -1)) dx \ \ is \ \ equivalent \ \ to \ \ (1)/(18) \int ( √(2)* sec^2 \theta - sec \theta) d\theta}`

Explanation:

`\int (x^3)/(√(9x^4+ 6x^2 -1)) dx = \int (x^3 \ dx)/(√((3x^22)^2 + 2(3x^2) (1) + 1-2))`

`\implies \int(x^3 \ dx)/(√((3x^2 +1)^2-2))`

let;

`3x^2 + 1 = √(2)\ sec \theta`

`6xdx = √(2) sec \theta tan \theta \ d \theta`

`xdx = (√(2))/(6) sec \theta tan \theta \ d \theta`

`\implies \int(x^2*x \ dx)/(√(2 \sec^2 \theta -2))`

`\implies \int((1)/(3)(√(2) \ sec \theta - 1)*(√(2))/(6) sec \theta tan \theta \ d\theta )/(√(2 ) \ tan \theta )`

`\implies (1)/(18) \int ( √(2)* sec \theta - 1) \ sec \theta d \theta`

`\implies (1)/(18) \int ( √(2)* sec^2 \theta - sec \theta) d\theta`

∴

`\mathbf{\int (x^3)/(√(9x^4+ 6x^2 -1)) dx \ \ is \ \ equivalent \ \ to \ \ (1)/(18) \int ( √(2)* sec^2 \theta - sec \theta) d\theta}`