Question

evaluate integral of 4y/✓(2y²+1).dy​

Answer 1

$\implies \: \displaystyle \int \bf \frac{4y}{2 {y}^{2} + 1 } dy$

$\implies \: \displaystyle \int \bf \frac{4y \: \: dy}{2 {y}^{2} + 1 }$

Now we will integrate the above expression by using substitution method.

$\bf \: let \: \: {2 {y}^{2} + 1 } = t$

Differentiating both sides :-

$\bf {4 {y} \: dy } = dt$

$\implies \: \displaystyle \int \bf \frac{dt}{t }$

We know that :-

$\underline{\boxed{\bf{ \displaystyle\int \: \frac{1}{x} \: dx = log(x) + c}}}$

Therefore :-

$\implies \: \displaystyle \int \bf \frac{dt}{t } = log(t) + c$

where c is constant.

Now put t = 2y²+1

$\implies \bf log(2 {y}^{2} + 1) + c$

Answer :

$\displaystyle \int \bf \frac{4y}{2 {y}^{2} + 1 } dy = log(2 {y}^{2} + 1) + c$

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Learn more :

∫ 1 dx = x + C

∫ sin x dx = – cos x + C

∫ cos x dx = sin x + C

∫ sec2 dx = tan x + C

∫ csc2 dx = -cot x + C

∫ sec x (tan x) dx = sec x + C

∫ csc x ( cot x) dx = – csc x + C

∫ (1/x) dx = ln |x| + C

∫ ex dx = ex+ C

∫ ax dx = (ax/ln a) + C

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