Explanation: 
Another Explanation (5):
সমাধান:
ধরি, \( I = \int_0^1 x^3 \sqrt{1+3x^4} \, dx \)
এখন, \( 1 + 3x^4 = u \) ধরি। সুতরাং, \( 12x^3 \, dx = du \) বা, \( x^3 \, dx = \frac{1}{12} du \)
যখন \( x = 0 \), তখন \( u = 1 + 3(0)^4 = 1 \)
যখন \( x = 1 \), তখন \( u = 1 + 3(1)^4 = 4 \)
অতএব,
\( I = \int_1^4 \sqrt{u} \cdot \frac{1}{12} \, du \)
\( = \frac{1}{12} \int_1^4 u^{\frac{1}{2}} \, du \)
\( = \frac{1}{12} \left[ \frac{u^{\frac{3}{2}}}{\frac{3}{2}} \right]_1^4 \)
\( = \frac{1}{12} \cdot \frac{2}{3} \left[ u^{\frac{3}{2}} \right]_1^4 \)
\( = \frac{1}{18} \left[ 4^{\frac{3}{2}} - 1^{\frac{3}{2}} \right] \)
\( = \frac{1}{18} \left[ (2^2)^{\frac{3}{2}} - 1 \right] \)
\( = \frac{1}{18} \left[ 2^3 - 1 \right] \)
\( = \frac{1}{18} \left[ 8 - 1 \right] \)
\( = \frac{1}{18} \cdot 7 \)
\( = \frac{7}{18} \) 🎉
সুতরাং, \( \int_0^1 x^3 \sqrt{1+3x^4} \, dx = \frac{7}{18} \) ✅
