We'll use the next basic integration formula:
∫ x n d x = x n + 1 n + 1 + C . \int x^ndx=\frac{x^{n+1}}{n+1}+C. ∫ x n d x = n + 1 x n + 1 + C .
So we have:
1)
∫ 2 x x d x = ∫ 2 x 1 x 1 2 d x = ∫ 2 x 3 2 d x = \int 2x\sqrt{x}dx=\int2x^1x^{\frac{1}{2}}dx=\int2x^{\frac{3}{2}}dx= ∫ 2 x x d x = ∫ 2 x 1 x 2 1 d x = ∫ 2 x 2 3 d x =
= 2 x 3 2 + 1 3 2 + 1 + C = 2 x 5 2 5 2 + C = 4 5 x 5 2 + C = =2\frac{x^{\frac{3}{2}+1}}{\frac{3}{2}+1}+C=2\frac{x^{\frac{5}{2}}}{\frac{5}{2}}+C=\frac{4}{5}x^{{\frac{5}{2}}}+C= = 2 2 3 + 1 x 2 3 + 1 + C = 2 2 5 x 2 5 + C = 5 4 x 2 5 + C =
= 4 5 x 2 + 1 2 + C = 4 5 x 2 x 1 2 + C = 4 5 x 2 x + C . =\frac{4}{5}x^{2+\frac{1}{2}}+C=\frac{4}{5}x^2x^{\frac{1}{2}}+C=\frac{4}{5}x^2\sqrt{x}+C. = 5 4 x 2 + 2 1 + C = 5 4 x 2 x 2 1 + C = 5 4 x 2 x + C .
2) ∫ 2 x 3 d x = ∫ 2 x 3 2 d x = ∫ 2 x − 3 2 d x = \int\frac{2}{\sqrt{x^3}}dx=\int\frac{2}{x^{\frac{3}{2}}}dx=\int2x^{-\frac{3}{2}}dx= ∫ x 3 2 d x = ∫ x 2 3 2 d x = ∫ 2 x − 2 3 d x =
= 2 x − 3 2 + 1 − 3 2 + 1 + C = 2 x − 1 2 − 1 2 + C = − 4 x 1 2 + C = − 4 x + C . =2\frac{x^{-\frac{3}{2}+1}}{-\frac{3}{2}+1}+C=2\frac{x^{-\frac{1}{2}}}{-\frac{1}{2}}+C=-\frac{4}{x^{\frac{1}{2}}}+C=-\frac{4}{\sqrt{x}}+C. = 2 − 2 3 + 1 x − 2 3 + 1 + C = 2 − 2 1 x − 2 1 + C = − x 2 1 4 + C = − x 4 + C .
Answer : 1) ∫ 2 x x d x = 4 5 x 2 x + C , \int 2x\sqrt{x}dx=\frac{4}{5}x^2\sqrt{x}+C, ∫ 2 x x d x = 5 4 x 2 x + C ,
2) ∫ 2 x 3 d x = − 4 x + C . \int\frac{2}{\sqrt{x^3}}dx=-\frac{4}{\sqrt{x}}+C. ∫ x 3 2 d x = − x 4 + C .
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