The value of displaystyle ∫ (d x/x (. xn + 1 .)) is equal to

Question

The value of displaystyle ∫ (d x/x (. xn + 1 .)) is equal to (A) (1/n) (log)e (. (xn/xn + 1) .) + c (B) (1/n) (log)e (. (xn + 1/xn) .) + c (C) (log)e (. (xn/xn + 1) .) + c (D) None of these

Tardigrade Admin · Accepted Answer

textI= displaystyle ∫ (d x/(. 1 + (1)xn .) × xn + 1) put 1 + (1/xn) = t - (n/xn + 1) . d x = d t I = displaystyle ∫ (- (d t/n)/t) = - (1/n) ln |. t |. + c I = - (1/n) ln (. 1 + (1/xn) .) + c I = (1/n) ln (. (xn/xn + 1) .) + c

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

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

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

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

None of these

Q. The value of ∫x(xn+1)dx​ is equal to

n1​(log)e​(xn+1xn​)+c

n1​(log)e​(xnxn+1​)+c

(log)e​(xn+1xn​)+c

None of these

I=∫(1+xn1​)×xn+1dx​ put 1+xn1​=t −xn+1n​.dx=dt I=∫t−ndt​​=−n1​ln∣t∣+c I=−n1​ln(1+xn1​)+c I=n1​ln(xn+1xn​)+c

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

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

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

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