Integrate the rational function in exercise: 1 x(x^n+1) [Hint: multiply numerator and denominator by x^ n – 1 and put x^n = t]

I= 1 x(x^n+1) dx Multiplying both numerator and denominator by nx^ n-1 , [ d dx (x^n+1)=nx^ n-1 ] I= nx^ n-1 nx^ n-1 .x(x^n+1) dx=1 n nx^ n-1 x^ n (x^n+1) dx Putting x^n = t nx^ n-1 = dx dt nx^ n-1 dx = dt From eq. (i), I=1 n dt t(t+1) =1 n 1 t(t+1) dt=1 n t+1-t t(t+1) dt =1 n t+1 t(t+1) - 1 t(t+1) dt=1 n [ 1 t dt+ 1 t+1 dt ] =1 n [log|t|-log|t+1| ]+c=1 n [log|x^n|-log|x^n+1| ]+c =1 n log | x^n x^n+1 |+c

Integrate the rational function in exercise: 1 x(x^n+1) [Hint: mu…

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Question

Integrate the rational function in exercise:
$\frac{1}{\text{x}(\text{x}^\text{n}+1)}$
[Hint: multiply numerator and denominator by $x^{n – 1}$ and put $x^n = t$]

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Answer

$\text{I}=\int\frac{1}{\text{x}(\text{x}^\text{n}+1)}\text{dx}$
Multiplying both numerator and denominator by $nx^{n-1}$,
$\Big[\therefore \ \frac{\text{d}}{\text{dx}}(\text{x}^\text{n}+1)=\text{nx}^{\text{n}-1}\Big]$
$\text{I}=\int\frac{\text{nx}^{\text{n}-1}}{\text{nx}^{\text{n}-1}.\text{x}(\text{x}^\text{n}+1)}\text{dx}=\frac{1}{\text{n}}\int\frac{\text{nx}^{\text{n}-1}}{\text{x}^{\text{n}}(\text{x}^\text{n}+1)}\text{dx}$
Putting $x^n = t$
$\Rightarrow nx^{n-1} = \frac{\text{dx}}{\text{dt}}$
$\Rightarrow nx^{n-1} dx = dt$
$\therefore$ From eq. (i),
$\text{I}=\frac{1}{\text{n}}\int\frac{\text{dt}}{\text{t}(\text{t}+1)}=\frac{1}{\text{n}}\int\frac{1}{\text{t}(\text{t}+1)}\text{dt}=\frac{1}{\text{n}}\int\frac{\text{t}+1-\text{t}}{\text{t}(\text{t}+1)}\text{dt}$
$=\frac{1}{\text{n}}\int\frac{\text{t}+1}{\text{t}(\text{t}+1)}-\int\frac{1}{\text{t}(\text{t}+1)}\text{dt}=\frac{1}{\text{n}}\Bigg[\int\frac{1}{\text{t}}\text{dt}+\int\frac{1}{\text{t}+1}\text{dt}\Bigg]$
$=\frac{1}{\text{n}}\Big[\text{log}|\text{t}|-\text{log}|\text{t}+1|\Big]+\text{c}=\frac{1}{\text{n}}\Big[\text{log}|\text{x}^\text{n}|-\text{log}|\text{x}^\text{n}+1|\Big]+\text{c}$
$=\frac{1}{\text{n}}\text{log}\Bigg|\frac{\text{x}^\text{n}}{\text{x}^\text{n}+1}\Bigg|+\text{c}$

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