+227 I get integrals and indefinite integrals. I do. I just don't get this one:
$$\int\frac{e^{sqrt(x)}}{sqrt(x)}$$
By the way, I couldn't figure out how to get square roots using LaTeX, so sqrt is square root.
Any help, please???
+33616 1. Use \sqrt{x} to get $$\sqrt{x}$$
2. Use the substitution:
$$u=\sqrt{x}$$
$$\\du=\frac{dx}{2\sqrt{x}}\\2\sqrt{x}du=dx\\2udu=dx$$
So:
$$\int \frac{e^{\sqrt{x}}}{\sqrt{x}}dx=\int \frac{e^{u} \times 2udu}{u}$$
or:
$$\int \frac{e^{\sqrt{x}}}{\sqrt{x}}dx=2\int e^{u}du = 2e^u=2e^{\sqrt{x}}$$
.
+33616 1. Use \sqrt{x} to get $$\sqrt{x}$$
2. Use the substitution:
$$u=\sqrt{x}$$
$$\\du=\frac{dx}{2\sqrt{x}}\\2\sqrt{x}du=dx\\2udu=dx$$
So:
$$\int \frac{e^{\sqrt{x}}}{\sqrt{x}}dx=\int \frac{e^{u} \times 2udu}{u}$$
or:
$$\int \frac{e^{\sqrt{x}}}{\sqrt{x}}dx=2\int e^{u}du = 2e^u=2e^{\sqrt{x}}$$
.
