∫ (dx/x√x6-16) = is equal to :

Question

∫ (dx/x√x6-16) = is equal to : (A) Sec-1((x3/4))+c  (B)  (1/12) Sec-1((x3/4))+c  (C) Cosh-1((x3/4))+c  (D)  (1/3) Sec-1((x3/4))+c

Tardigrade Admin · Accepted Answer

Let I=∫ (d x/x √x6-16) =(1/3) ∫ (3 x2/x3 √(x3)2-42) d x Put x3 =t ⇒ 3 x2 d x =d t ∴ I =(1/3) ∫ (d t/t √t2-42) =(1/3 × 4) sec -1((t/4))+c =(1/12) sec -1((x3/4))+c

Q. $\int \frac{d x}{x x ^{6} - 16}$ = is equal to :

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

$\frac{1}{12} S e c^{- 1} (\frac{x ^{3}}{4}) + c$

$C os h^{- 1} (\frac{x ^{3}}{4}) + c$

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

Q. ∫xx6−16​dx​= is equal to :

Sec−1(4x3​)+c

121​Sec−1(4x3​)+c

Cosh−1(4x3​)+c

31​Sec−1(4x3​)+c

Let I=∫xx6−16​dx​ =31​∫x3(x3)2−42​3x2​dx Put x3=t ⇒3x2dx=dt ∴I=31​∫tt2−42​dt​ =3×41​sec−1(4t​)+c =121​sec−1(4x3​)+c

Q. $\int \frac{d x}{x x ^{6} - 16}$ = is equal to :

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

$\frac{1}{12} S e c^{- 1} (\frac{x ^{3}}{4}) + c$

$C os h^{- 1} (\frac{x ^{3}}{4}) + c$

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