shamieh
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Suppose:
$$\int^{\infty}_1 \frac{x}{x^2 + 1} \, dx$$
I can't do this can I?
Let $$u = x^2 + 1$$
so $$du = 2x \, dx$$
$$\therefore \frac{du}{2} = x \, dx$$ ?
$$\int^{\infty}_1 \frac{x}{x^2 + 1} \, dx$$
I can't do this can I?
Let $$u = x^2 + 1$$
so $$du = 2x \, dx$$
$$\therefore \frac{du}{2} = x \, dx$$ ?