If ∫ (1/x+x5)dx =f(x)+c , then ∫ (x4/x+x5) dx =

Question

If ∫ (1/x+x5)dx =f(x)+c , then ∫ (x4/x+x5) dx =  (A)  log | x | + f(x) + c  (B)  log | x | - f(x) + c  (C)  log | x | + x f(x) + c  (D)  log | x | - x f(x) + c

Tardigrade Admin · Accepted Answer

∫ (1/x+x5)dx =f(x)+c I = ∫ (x4/x+x5) dx = ∫ (x4 +1 -1/x (1+x4)) dx = ∫ (x4+1/x(1+x4)) dx - ∫ (1/x(1+x4)) dx = ∫ (1/x) dx - ∫ (1/x+x5) dx = log |x| - f(x) + c

Q. If $\int \frac{1}{x + x ^{5}} d x = f (x) + c,$ then $\int \frac{x ^{4}}{x + x ^{5}} d x =$

$lo g ∣ x ∣ + f (x) + c$

$lo g ∣ x ∣ - f (x) + c$

$lo g ∣ x ∣ + x f (x) + c$

$lo g ∣ x ∣ - x f (x) + c$

Q. If ∫x+x51​dx=f(x)+c, then ∫x+x5x4​dx=

log∣x∣+f(x)+c

log∣x∣−f(x)+c

log∣x∣+xf(x)+c

log∣x∣−xf(x)+c

∫x+x51​dx=f(x)+c I=∫x+x5x4​dx=∫x(1+x4)x4+1−1​dx =∫x(1+x4)x4+1​dx−∫x(1+x4)1​dx =∫x1​dx−∫x+x51​dx =log∣x∣−f(x)+c

Q. If $\int \frac{1}{x + x ^{5}} d x = f (x) + c,$ then $\int \frac{x ^{4}}{x + x ^{5}} d x =$

$lo g ∣ x ∣ + f (x) + c$

$lo g ∣ x ∣ - f (x) + c$

$lo g ∣ x ∣ + x f (x) + c$

$lo g ∣ x ∣ - x f (x) + c$