Question

Nesecito ayuda es para alrrato que estén resueltas a todo detalle xfa si no reprobar calculo :(​

Answer 1

Asi quedarían tus integrales:

1)

$\int \sqrt{2x+1}dx= \int \sqrt{u} \frac{du}{2}= \frac{1}{2} \int u^{1/2}du= \frac{1}{2} (u^{3/2} \frac{2}{3})+C= \frac{(2x+1)^{3/2}}{3}+C \\\\u=2x+1\\\\du=2dx--- > dx=\frac{du}{2}$

2)

$\int \frac{dx}{(2x+1)^2}= \int \frac{du}{2u^2}= \frac{1}{2} \int u^{-2}du= \frac{1}{2} (-u^{-1})+C=- \frac{1}{2(2x+1)} +C \\\\u=2x+1\\\\du=2dx--- > dx=\frac{du}{2}$

3)

$\int 6x^2(3x^3+2)dx= \int (18x^5+12x^2)dx =18 \frac{x^6}{6} +12 \frac{x^3}{3}+C=3x^6+4x^3+C$

4)

$\int 2e^{5x}dx=2 \int e^u \frac{du}{5} = \frac{2}{5} e^u +C= \frac{2}{5} e^{5x} +C \\\\u=5x\\\\du=5dx--- > dx=\frac{du}{5}$

5)

$\int xe^{x^2}dx= \int e^u \frac{du}{2}= \frac{1}{2} e^u +C = \frac{1}{2} e^{x^2} +C \\\\u=x^2\\\\du=2xdx--- > xdx=\frac{du}{2}$

6)

$\int \frac{2dx}{6x} =\frac{1}{3} \int \frac{dx}{x} = \frac{1}{3} Ln(x)+C$

7)

$\int \frac{3x^2dx}{4x^3} = \frac{3}{4} \int \frac{dx}{x} = \frac{3}{4} Ln(x)+C$