∫(dx/x(xn+1))is equal to

Question

∫(dx/x(xn+1))is equal to (A) (1/n) log ((xn/xn+1))+c (B) (1/n) log ((xn+1/xn))+c (C)  log ((xn/xn+1))+c (D) none of these

Tardigrade Admin · Accepted Answer

I=∫(dx/x(xn+1))=∫(xn-1/x2(xn+1) )dx Putting xn = t so that nxn-1dx = dt, i.e. we get xn-1dx = (1/n)dt I = ∫((1/n)dt/t(t-1)) = (1/n)∫((1/t)-(1/t+1))dt =(1/n)(log t-log (t+1))+C =(1/n) log ((xn/xn+1))+C

Q. $\int \frac{d x}{x ( x ^{n} + 1 )}$ is equal to

$\frac{1}{n} l o g (\frac{x ^{n}}{x ^{n} + 1}) + c$

$\frac{1}{n} l o g (\frac{x ^{n} + 1}{x ^{n}}) + c$

$l o g (\frac{x ^{n}}{x ^{n} + 1}) + c$

none of these

Q. ∫x(xn+1)dx​is equal to

n1​log(xn+1xn​)+c

n1​log(xnxn+1​)+c

log(xn+1xn​)+c

none of these

I=∫x(xn+1)dx​=∫x2(xn+1)xn−1​dx Putting xn=t so that nxn−1dx=dt,i.e. we get xn−1dx=n1​dt I=∫t(t−1)n1​dt​=n1​∫(t1​−t+11​)dt =n1​(logt−log(t+1))+C =n1​log(xn+1xn​)+C

Q. $\int \frac{d x}{x ( x ^{n} + 1 )}$ is equal to

$\frac{1}{n} l o g (\frac{x ^{n}}{x ^{n} + 1}) + c$

$\frac{1}{n} l o g (\frac{x ^{n} + 1}{x ^{n}}) + c$

$l o g (\frac{x ^{n}}{x ^{n} + 1}) + c$