Question

$\huge \displaystyle \sf\int \limits_{ \sqrt[3]{2} }^{ \infty } \frac{1}{x \sqrt{ {x}^{3} - 1} } \: dx$​

Answer 1

Answer:

$\huge \red{ \frac{\pi}{6} }$

Step-by-step explanation:

$\displaystyle \sf\int \limits_{ \sqrt[3]{2} }^{ \infty } \frac{1}{x \sqrt{ {x}^{3} - 1} } \: dx \\ \\ = \displaystyle \sf\int \limits_{ \sqrt[3]{2} }^{ \infty } \frac{ {x}^{2} }{x {}^{3} \sqrt{ {x}^{3} - 1} } \: dx\\ \\ put \: {x}^{3} - 1 = y \implies \: 3 {x}^{2} dx = dy \\ \\ \displaystyle \sf\int \limits_{ 1}^{ \infty } \frac{1}{3(1 + y) \sqrt{ y} } \: dy \\ \\ put \: y = \tan {}^{2} ( \theta) \implies \: dy = 2 \tan( \theta) \sec {}^{2} ( \theta) d\theta \\ \\ \displaystyle \sf\int \limits_{ \frac{\pi}{4} }^{ \frac{\pi}{2} } \frac{ 2 \tan( \theta) \sec {}^{2} ( \theta)}{ 3(1 + \tan {}^{2} ( \theta) ) \sqrt{ \tan {}^{2} ( \theta) } }d \theta \\ \\ \displaystyle \sf\int \limits_{ \frac{\pi}{4} }^{ \frac{\pi}{2} } \frac{ 2 \tan( \theta) \sec {}^{2} ( \theta)}{3\tan( \theta) \sec {}^{2}( \theta)}d \theta \\ \\ = \frac{2}{3} \bigg[ \theta \bigg] _{ \frac{\pi}{4} } ^{ \frac{\pi}{2} } \\ \\ = \frac{2}{3} ( \frac{\pi}{2} - { \frac{\pi}{4} }) \\ \\ = \frac{2}{3} \times \frac{\pi}{4} = \frac{\pi}{6}$

Answer 2

$\large\underline{\underline{\text{Question:}}}$

• $\sf\int \limits_{ \sqrt[3]{2} }^{ \infty } \frac{1}{x \sqrt{ {x}^{3} - 1} } \: dx$

$\large\underline{\underline{\text{Solution:}}}$

$\longrightarrow I = \int \limits_{ \sqrt[3]{2} }^{ \infty } \frac{1}{x \sqrt{ {x}^{3} - 1} } \: dx$

$\longrightarrow I = \int \limits_{ \sqrt[3]{2} }^{ \infty } \frac{1}{x} \times \frac{1}{ \sqrt{ {x}^{3} - 1} } \: dx$

Multiple numerator and denominator by x²,

$\longrightarrow I = \int \limits_{ \sqrt[3]{2} }^{ \infty } \frac{ {x}^{2} }{ {x}^{2} } \times \frac{1}{x} \times \frac{1}{ \sqrt{ {x}^{3} - 1} } \: dx$

$\longrightarrow I = \int \limits_{ \sqrt[3]{2} }^{ \infty } \frac{ {x}^{2} }{ {x}^{3} } \times \frac{1}{ \sqrt{ {x}^{3} - 1} } \: dx$

$\longrightarrow I = \int \limits_{ \sqrt[3]{2} }^{ \infty } \frac{ {x}^{2} }{3 {x}^{3} } \times \frac{1}{ \sqrt{ {x}^{3} - 1} } \: dx$

Let's,

• ${x}^{3} - 1 = y$

So,

• ${x}^{3} = y + 1$

$\longrightarrow I = \int \limits_{ \sqrt[3]{2} }^{ \infty } \frac{ {x}^{2} }{3(y + 1)} \times \frac{1}{ \sqrt{y} } \: dx$

$\longrightarrow I = \int \limits_{ 1}^{ \infty } \frac{1}{ 3(y + 1)\sqrt{y} } \: dy$

Let's,

• $y = \tan^{2} ( \theta)$

So,

$\longrightarrow I = \int \limits_{ 1 }^{ \infty } \frac{2 \tan( \theta) { \sec }^{2}( \theta) }{ 3( \tan^{2} ( \theta) + 1)\sqrt{ {\tan}^{2} ( \theta)} } \: d \theta$

As we know that,

• $\tan( \frac{\pi}{4} ) = 1$
• $\tan( \frac{\pi}{2} ) = \infty$

$\longrightarrow I = \int \limits_{ \frac{\pi}{4} }^{ \frac{\pi}{2} } \frac{2 \tan ( \theta) { \sec }^{2}( \theta) }{ 3( \tan^{2} ( \theta) + 1)\sqrt{ \tan ^{2} ( \theta)} } \: d \theta$

$\longrightarrow I = \int \limits_{ \frac{\pi}{4} }^{ \frac{\pi}{2} } \frac{2 \tan ( \theta) { \sec }^{2}( \theta) }{ 3 \sec^{2} ( \theta) \tan( \theta) } \: d \theta \: \: \: \: \: \: \: \: [1 + {\tan}^{2} ( \theta) = \sec ^{2}( \theta) ]$

$\longrightarrow I = \frac{2}{3} \bigg( \theta \bigg)_ { \frac{\pi}{4} }^{ \frac{\pi}{2} }$

$\longrightarrow I = \frac{2}{3} \bigg( \frac{\pi}{2} - \frac{\pi}{4} \bigg)$

$\longrightarrow I = \frac{2}{3} \bigg( \frac{2\pi - \pi}{4} \bigg)$

$\longrightarrow I = \frac{2}{3} \bigg( \frac{\pi}{4} \bigg)$

$\longrightarrow I = \frac{\pi}{6}$

$\large\underline{\underline{\text{Final Answer:}}}$

• $\longrightarrow \boxed {\int \limits_{ \sqrt[3]{2} }^{ \infty } \frac{1}{x \sqrt{ {x}^{3} - 1} } \: dx = \frac{\pi}{6} }$

❥Hope you get your AnSwEr.