4.
[tex]\(\int \frac{(1-\sqrt{x})^2}{\sqrt{x}} \, dx\)[/tex]
5. [tex]\(\int \left(\frac{3}{\sqrt[3]{x}} - \sqrt{x} + \pi\right) \, dx\)[/tex]
6. [tex]\(\int \frac{x^2 + 5x + 6}{x+3} \, dx\)[/tex]
7. [tex]\(\int \frac{1}{x} \, dx\)[/tex]
8. [tex]\(\int \frac{1}{3x^3} \, dx\)[/tex]
Answer (1)
Let's solve each integral one by one:
∫ x ( 1 − x ) 2 d x First, expand ( 1 − x ) 2 to get 1 − 2 x + x . Then rewrite the integral as: ∫ ( x 1 − 2 + x ) d x . Separating the integrals, we have: ∫ x 1 d x − 2 ∫ 1 d x + ∫ x d x . Solve each part separately:
∫ x 1 d x = ∫ x − 1/2 d x = 2 x 1/2 + C 1
− 2 ∫ 1 d x = − 2 x + C 2
∫ x d x = ∫ x 1/2 d x = 3 2 x 3/2 + C 3 Combine the results: 2 x 1/2 − 2 x + 3 2 x 3/2 + C .
∫ ( 3 x 3 − x + π ) d x Rewrite the integral by expressing each term with exponents: ∫ 3 x − 1/3 d x − ∫ x 1/2 d x + ∫ π d x . Solve each part separately:
∫ 3 x − 1/3 d x = 3 ⋅ 2 3 x 2/3 + C 1 = 2 9 x 2/3 + C 1
∫ x 1/2 d x = 3 2 x 3/2 + C 2
∫ π d x = π x + C 3 Combine the results: 2 9 x 2/3 − 3 2 x 3/2 + π x + C .
∫ x + 3 x 2 + 5 x + 6 d x Perform polynomial long division on x 2 + 5 x + 6 by x + 3 .
Dividing, we get a quotient of x + 2 and a remainder of 0 . Therefore, the integral reduces to: ∫ ( x + 2 ) d x . Solve: 2 x 2 + 2 x + C .
∫ x 1 d x This is a standard integral that results in: ln ∣ x ∣ + C .
∫ 3 x 3 1 d x Rewrite as 3 1 ∫ x − 3 d x , which evaluates to: 3 1 ⋅ ( − 2 1 x − 2 ) + C = − 6 x 2 1 + C .
