Question-164376

Question Number 164376 by muneer0o0 last updated on 16/Jan/22

Answered by alephzero last updated on 16/Jan/22

\int \frac{d x}{\sqrt{x^{3} + 64}} = ?

x^{3} = {(\sqrt{x^{3}})}^{2}

\int \frac{d u}{\sqrt{u^{2} + a}} = \ln ∣ u + \sqrt{u^{2} + a} ∣ + C

\Rightarrow \int \frac{d x}{\sqrt{{(\sqrt{x^{3}})}^{2} + 64}} = \ln ∣ \sqrt{x^{3}} + \sqrt{x^{3} + 64} ∣ + C

I think I wrong .

(I just started learn calculus) .

Commented by mr W last updated on 16/Jan/22

y e s, v e r y w r o n g! {i t}^{'} s e v e n n o t i n t e g r a b l e

u s i n g e l e m e n t a r y f u n c t i o n s .

Answered by ajfour last updated on 16/Jan/22

I = \int \frac{d x}{{(x^{3} + a^{3})}^{1 / 2}}

l e t x = a {(\tan^{2} θ)}^{1 / 3}

d x = \frac{2 (1 + \tan^{2} θ) d θ}{3 {(\tan θ)}^{1 / 3}}

I = \int \frac{2 d θ}{3 a \sqrt{a} {(\tan θ)}^{1 / 3} \cos θ}

= \frac{2}{3 a \sqrt{a}} \int \frac{d θ}{\sin^{1 / 3} θ \cos^{2 / 3} θ}

I t h i n k n o w G a m m a f u n c t i o n

s h o u l d b e u s e d \dots

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