Solveeit Logo
Login

Question

Question: \[\int_{}^{}{\frac{\mathbf{dx}}{\mathbf{x(}\mathbf{x}^{\mathbf{n}}\mathbf{+ 1)}}\mathbf{=}}\]...

∫dxx(xn+1)=\int_{}^{}{\frac{\mathbf{dx}}{\mathbf{x(}\mathbf{x}^{\mathbf{n}}\mathbf{+ 1)}}\mathbf{=}}

A

1nlog⁡xnxn+1+c\frac{1}{n}\log\frac{x^{n}}{x^{n} + 1} + c

B

nlog⁡xn+1xn+cn\log\frac{x^{n} + 1}{x^{n}} + c

C

−1nlog⁡xnxn+1+c\frac{- 1}{n}\log\frac{x^{n}}{x^{n} + 1} + c

D

−nlog⁡xn+1xn+c- n\log\frac{x^{n} + 1}{x^{n}} + c

Answer

1nlog⁡xnxn+1+c\frac{1}{n}\log\frac{x^{n}}{x^{n} + 1} + c

Explanation

Solution

Let I= ∫dxx(xn+1)=∫xn−1xn(xn+1)dxI = \ \int_{}^{}{\frac{dx}{x(x^{n} + 1)} = \int_{}^{}\frac{x^{n - 1}}{x^{n}(x^{n} + 1)}dx}

Putting xn=t⇒nxn−1dx=dt,x^{n} = t \Rightarrow nx^{n - 1}dx = dt, we have

I=1n∫dtt(t+1)=1n∫[1t−1t+1]dt,I = \frac{1}{n}\int_{}^{}{\frac{dt}{t(t + 1)} = \frac{1}{n}\int_{}^{}\left\lbrack \frac{1}{t} - \frac{1}{t + 1} \right\rbrack}dt, (by resolving into partial fractions)

=1n[log⁡t−log⁡(t+1)]+c= \frac{1}{n}\lbrack\log t - \log(t + 1)\rbrack + c =1nlog⁡∣tt+1∣+c=1nlog⁡∣xnxn+1∣+c.= \frac{1}{n}\log\left| \frac{t}{t + 1} \right| + c = \frac{1}{n}\log\left| \frac{x^{n}}{x^{n} + 1} \right| + c.