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levacccp [35]
3 years ago
15

Which of these integrals is equivalent to the given integral?

Mathematics
1 answer:
ladessa [460]3 years ago
6 0

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}

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