∫_0 ^x (t^2 /( (√(a+2t^2 ))))dt


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Question Number 166141 by Lambert last updated on 13/Feb/22

$\int_{0}^{x} \frac{t^{2}}{\sqrt{a + 2 t^{2}}} dt$
	Answered by MJS_new last updated on 14/Feb/22

	$\int \frac{t^{2}}{\sqrt{a + 2 t^{2}}} d t =$ $[u = \frac{\sqrt{2}}{\sqrt{a}} (t + \sqrt{a + 2 t^{2}}) \to d t = \frac{\sqrt{a + 2 t^{2}}}{\sqrt{2} u} d u]$ $= \frac{a \sqrt{2}}{16} \int \frac{{(u^{2} - 1)}^{2}}{u^{3}} d u = \frac{a \sqrt{2}}{16} \int (u - \frac{2}{u} + \frac{1}{u^{3}}) d u =$ $= \frac{a \sqrt{2}}{16} (\frac{u^{2}}{2} - 2 \ln u - \frac{1}{2 u^{2}}) = . . .$ $= \frac{t \sqrt{a + 2 t^{2}}}{4} - \frac{a \sqrt{2}}{8} \ln (\sqrt{2} t + \sqrt{a + 2 t^{2}}) + C$ $the rest is easy$
	Commented by cortano1 last updated on 14/Feb/22

	$u = \frac{\sqrt{2}}{\sqrt{a}} t ? sir$
	Commented by MJS_new last updated on 14/Feb/22

	$typo = instead of +; {it}^{'} s now corrected$
	Answered by Mathspace last updated on 14/Feb/22
	$f(a)=(1/( (√a)))∫_0 ^x (t^2 /( (√(1+((2t^2 )/a)))))dt =(a/2)×(1/( (√a)))∫_0 ^x (((2/a)t^2 )/( (√(1+((2t^2 )/a)))))dt =((√a)/2)∫_0 ^x (√(1+((2t^2 )/a)))dt−((√a)/2)∫_0 ^x (dt/( (√(1+((2t^2 )/a))))) (((√2)t)/( (√a)))=shu ⇒u=argsh((((√2)t)/( (√a)))) ∫_0 ^x (√(1+((2t^2 )/a)))dt=∫_0 ^(argsh((((√2)x)/( (√a))))) chu.((√a)/( (√2)))chudu =((√a)/( (√2)))∫_0 ^(ln((((√2)x)/( (√a)))+(√(1+((2x^2 )/a))))) ((1+ch(2u))/2)du =((√a)/2)ln(....)+((√a)/(4(√2)))[sh(2u)]_0 ^(ln(...)) =((√a)/2)ln(...)+((√a)/(4(√2)))[((e^(2u) −e^(−2u) )/2)]_0 ^(ln(...)) =((√a)/2)ln((((√2)x)/( (√a)))+(√(1+((2x^2 )/a)))) ((√a)/(8(√2))){ ((((√2)x)/( (√a)))+(√(1+((2x^2 )/a))))^2 −((((√2)x)/( (√a)))+(√(1+((2x^2 )/a))))^(−2) } same way for other integral...$
	$f (a) = \frac{1}{\sqrt{a}} \int_{0}^{x} \frac{t^{2}}{\sqrt{1 + \frac{2 t^{2}}{a}}} d t$ $= \frac{a}{2} \times \frac{1}{\sqrt{a}} \int_{0}^{x} \frac{\frac{2}{a} t^{2}}{\sqrt{1 + \frac{2 t^{2}}{a}}} d t$ $= \frac{\sqrt{a}}{2} \int_{0}^{x} \sqrt{1 + \frac{2 t^{2}}{a}} d t - \frac{\sqrt{a}}{2} \int_{0}^{x} \frac{d t}{\sqrt{1 + \frac{2 t^{2}}{a}}}$ $\frac{\sqrt{2} t}{\sqrt{a}} = s h u \Rightarrow u = a r g s h (\frac{\sqrt{2} t}{\sqrt{a}})$ $\int_{0}^{x} \sqrt{1 + \frac{2 t^{2}}{a}} d t = \int_{0}^{a r g s h (\frac{\sqrt{2} x}{\sqrt{a}})} c h u . \frac{\sqrt{a}}{\sqrt{2}} c h u d u$ $= \frac{\sqrt{a}}{\sqrt{2}} \int_{0}^{l n (\frac{\sqrt{2} x}{\sqrt{a}} + \sqrt{1 + \frac{2 x^{2}}{a}})} \frac{1 + c h (2 u)}{2} d u$ $= \frac{\sqrt{a}}{2} l n (. . . .) + \frac{\sqrt{a}}{4 \sqrt{2}} {[s h (2 u)]}_{0}^{l n (. . .)}$ $= \frac{\sqrt{a}}{2} l n (. . .) + \frac{\sqrt{a}}{4 \sqrt{2}} {[\frac{e^{2 u} - e^{- 2 u}}{2}]}_{0}^{l n (. . .)}$ $= \frac{\sqrt{a}}{2} l n (\frac{\sqrt{2} x}{\sqrt{a}} + \sqrt{1 + \frac{2 x^{2}}{a}})$ $\frac{\sqrt{a}}{8 \sqrt{2}} {{(\frac{\sqrt{2} x}{\sqrt{a}} + \sqrt{1 + \frac{2 x^{2}}{a}})}^{2} - {(\frac{\sqrt{2} x}{\sqrt{a}} + \sqrt{1 + \frac{2 x^{2}}{a}})}^{- 2}}$ $s a m e w a y f o r o t h e r i n t e g r a l . . .$
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