∫ ( x 3 + 1 ) 5 x 2 d x \int \left(x^3+1\right)^5x^2dx ∫ ( x 3 + 1 ) 5 x 2 d x
Here, we apply substitution method I.e.
Let u = x 3 + 1 u = x^3 + 1 u = x 3 + 1
⟹ d u d x = 3 x 2 \implies \frac{du}{dx} = 3x^2 ⟹ d x d u = 3 x 2
⟹ d x = 1 3 x 2 d u \implies dx= \frac{1}{3x^2}du ⟹ d x = 3 x 2 1 d u
Replacing back, we have;
∫ ( u ) 5 x 2 1 3 x 2 d u \int (u)^5 \cancel{x^2} \frac{1}{3 \cancel{x^2} }du ∫ ( u ) 5 x 2 3 x 2 1 d u = 1 3 ∫ u 5 d u = \frac{1}{3} \int u^5 du = 3 1 ∫ u 5 d u
Now we solve;
∫ u 5 d u \int u^5 du ∫ u 5 d u
Here, we apply power rule;
Where we let;
∫ u n d u = u n + 1 n + 1 \int u^n du = \frac{u^{n+1}}{n+1} ∫ u n d u = n + 1 u n + 1 With n = 5 n = 5 n = 5 , we have;
= u 6 6 = \frac{u^6}{6} = 6 u 6
Replacing this back to 1 3 ∫ u 5 d u \frac{1}{3} \int u^5 du 3 1 ∫ u 5 d u we have;
u 6 18 \frac{u^6}{18} 18 u 6
Now we undo the substitution
u = x 3 + 1 u = x^3 +1 u = x 3 + 1
( x 3 + 1 ) 6 18 \frac{(x^3 +1)^6}{18} 18 ( x 3 + 1 ) 6 and we add a C to this, hence the final answer i.e.
( x 3 + 1 ) 6 18 + C \frac{(x^3 +1)^6}{18} + C 18 ( x 3 + 1 ) 6 + C