Question

Which of these integrals is equivalent to the given integral?

Answer 1

Answer:

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

Step-by-step explanation:

$\int \dfrac{x^3}{\sqrt{9x^4+ 6x^2 -1}} dx = \int \dfrac{x^3 \ dx}{\sqrt{(3x^22)^2 + 2(3x^2) (1) + 1-2}}$

$\implies \int\dfrac{x^3 \ dx}{\sqrt{(3x^2 +1)^2-2}}$

let;

$3x^2 + 1 = \sqrt{2}\ sec \theta$

$6xdx = \sqrt{2} sec \theta tan \theta \ d \theta$

$xdx = \dfrac{\sqrt{2}}{6} sec \theta tan \theta \ d \theta$

$\implies \int\dfrac{x^2*x \ dx}{\sqrt{2 \sec^2 \theta -2}}$

$\implies \int\dfrac{\dfrac{1}{3}(\sqrt{2} \ sec \theta - 1)*\dfrac{\sqrt{2}}{6} sec \theta tan \theta \ d\theta }{\sqrt{2 } \ tan \theta }$

$\implies \dfrac{1}{18} \int ( \sqrt{2}* sec \theta - 1) \ sec \theta d \theta$

$\implies \dfrac{1}{18} \int ( \sqrt{2}* sec^2 \theta - sec \theta) d\theta$

∴

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