If I =∫(x5/√1+x3)dx , then I is equal to

Question

If I =∫(x5/√1+x3)dx , then I is equal to (A) (2/9)(1+x3)(5/2)+(2/3)(1+x3)(3/2)+C (B) log|√x+√1+x3|+C (C) log|√x-√1+x3|+C (D) (2/9)(1+x3)(3/2)-(2/3)(1+x3)(1/2)+C

Tardigrade Admin · Accepted Answer

I =∫(x5/√1+x3)dx=∫ (x3.x2/√1.+x3)dx Let 1 + x3 = t2, so that 3x2 dx = 2t dt ⇒ x2 dx = (2/3)td t ∴ I =∫((t2-1)(2/3)t dt/t)=(2/3)∫ t2 -1)dt =(2/3)((t3/3)-t)+C =(2/3)[((1+x3)3/2/3)-(1+x3)(1/2)]+C =(2/9)(1+x3)3/2-(2/3)(1+x3)1/2+C

Q. If $I = \int \frac{x ^{5}}{1 + x ^{3}} d x$ , then I is equal to

$\frac{2}{9} (1 + x^{3})^{\frac{5}{2}} + \frac{2}{3} (1 + x^{3})^{\frac{3}{2}} + C$

$l o g ∣ ∣ x + 1 + x^{3} ∣ ∣ + C$

$l o g ∣ ∣ x - 1 + x^{3} ∣ ∣ + C$

$\frac{2}{9} (1 + x^{3})^{\frac{3}{2}} - \frac{2}{3} (1 + x^{3})^{\frac{1}{2}} + C$

Q. If I=∫1+x3​x5​dx , then I is equal to

92​(1+x3)25​+32​(1+x3)23​+C

log∣∣​x​+1+x3​∣∣​+C

log∣∣​x​−1+x3​∣∣​+C

92​(1+x3)23​−32​(1+x3)21​+C

I=∫1+x3​x5​dx=∫1.+x3​x3.x2​dx Let1+x3=t2,sothat3x2dx=2tdt ⇒x2dx=32​tdt ∴I=∫t(t2−1)32​tdt​=32​∫t2−1)dt =32​(3t3​−t)+C=32​[3(1+x3)3/2​−(1+x3)21​]+C =92​(1+x3)3/2−32​(1+x3)1/2+C

Q. If $I = \int \frac{x ^{5}}{1 + x ^{3}} d x$ , then I is equal to

$\frac{2}{9} (1 + x^{3})^{\frac{5}{2}} + \frac{2}{3} (1 + x^{3})^{\frac{3}{2}} + C$

$l o g ∣ ∣ x + 1 + x^{3} ∣ ∣ + C$

$l o g ∣ ∣ x - 1 + x^{3} ∣ ∣ + C$

$\frac{2}{9} (1 + x^{3})^{\frac{3}{2}} - \frac{2}{3} (1 + x^{3})^{\frac{1}{2}} + C$