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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

Sagot :

ajeigbeibraheem

Answer:

[tex]\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}[/tex]

Step-by-step explanation:

[tex]\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}}[/tex]

[tex]\implies \int\dfrac{x^3 \ dx}{\sqrt{(3x^2 +1)^2-2}}[/tex]

let;

[tex]3x^2 + 1 = \sqrt{2}\ sec \theta[/tex]

[tex]6xdx = \sqrt{2} sec \theta tan \theta \ d \theta[/tex]

[tex]xdx = \dfrac{\sqrt{2}}{6} sec \theta tan \theta \ d \theta[/tex]

[tex]\implies \int\dfrac{x^2*x \ dx}{\sqrt{2 \sec^2 \theta -2}}[/tex]

[tex]\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 }[/tex]

[tex]\implies \dfrac{1}{18} \int ( \sqrt{2}* sec \theta - 1) \ sec \theta d \theta[/tex]

[tex]\implies \dfrac{1}{18} \int ( \sqrt{2}* sec^2 \theta - sec \theta) d\theta[/tex]

∴

[tex]\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}[/tex]

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