∫ ((x3-1)/(x4+1)(x+1)) d x=

Question

∫ ((x3-1)/(x4+1)(x+1)) d x= (A) (1/4) ln (1+x4)+(1/3) ln (1+x3)+c (B) (1/4) ln (1+x4)-(1/3) ln (1+x3)+c (C) (1/4) ln (1+x4)- ln (1+x)+c (D) (1/4) ln (1+x4)+ ln (1+x)+c

Tardigrade Admin · Accepted Answer

∫ (x3-1/(x4+1)(x+1)) d x =∫ ((x4+x3)-(x4+1)/(x4+1)(x+1)) d x =∫ (x3/x4+1) d x-∫ (1/x+1) d x =(1/4) ln (x4+1)- ln (x+1)+c

Q. $\int \frac{( x ^{3} - 1 )}{( x ^{4} + 1 ) ( x + 1 )} d x =$

$\frac{1}{4} ln (1 + x^{4}) + \frac{1}{3} ln (1 + x^{3}) + c$

$\frac{1}{4} ln (1 + x^{4}) - \frac{1}{3} ln (1 + x^{3}) + c$

$\frac{1}{4} ln (1 + x^{4}) - ln (1 + x) + c$

$\frac{1}{4} ln (1 + x^{4}) + ln (1 + x) + c$

$\int \frac{x ^{3} - 1}{( x ^{4} + 1 ) ( x + 1 )} d x = \int \frac{( x ^{4} + x ^{3} ) - ( x ^{4} + 1 )}{( x ^{4} + 1 ) ( x + 1 )} d x$
$= \int \frac{x ^{3}}{x ^{4} + 1} d x - \int \frac{1}{x + 1} d x$
$= \frac{1}{4} ln (x^{4} + 1) - ln (x + 1) + c$

Q. ∫(x4+1)(x+1)(x3−1)​dx=

41​ln(1+x4)+31​ln(1+x3)+c

41​ln(1+x4)−31​ln(1+x3)+c

41​ln(1+x4)−ln(1+x)+c

41​ln(1+x4)+ln(1+x)+c