Question

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integral : 1/(4x^3+3x^2+2x+1)......​

Answer 1

To Find:

∫ [1/(4x³+3x²+2x+1)] dx

♦Solution♦

Now, let; u=4x³+3x²+2x+1

» y=1/u

On applying chain rule:

, i.e., ∫(y) dx =∫(y) du ÷ [(d/dx){u}]

∫(y) dx=∫(1/u) du÷[(d/dx){4x³+3x²+2x+1}]

$⇒ \frac{ log_{e}(u) }{12x {}^{2} + 6x + 2 } + C \: \: \: , \ \: \: C\: ∈ \: R$

$⇒ \frac{ln(4 {x}^{3} + 3 {x}^{2} + 2x + 1 )}{12x {}^{2} + 6x + 2} + C \:, \: C\: ∈ \: R$

Answer 2

Explanation:

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