How do I evaluate $\int \frac{3 x}{x^{2} + 1} d x$ $int(3x)/(x^2+1)dx$ ?

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How do I evaluate $\int \frac{3 x}{x^{2} + 1} d x$ $int(3x)/(x^2+1)dx$ ?

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Answer 1 · 2015-03-05T21:53:40

Use substitution (often called $u$ $u$ substitution).

Notice that the derivative of the denominator is $2 x$ $2x$ , which is a lot like what's in the numerator. The $3$ $3$ in the numerator is kinda in our way, so move it outside the integral sign.

(Recall that $\int c f (x) d x = c \int f (x) d x$ $intcf(x)dx=cintf(x)dx$ , so this wont't change the integral.)

Now the integral is $3 \int \frac{x}{x^{2} + 1} d x$ $3intx/(x^2+1)dx$ .

Let $u = x^{2} + 1$ $u=x^2+1$ making $d u = 2 x d x$ $du=2xdx$ and proceed with whatever details of substitution you've learned.

I use: $3 \int \frac{x}{x^{2} + 1} d x = 3 \int \frac{1}{2} \frac{2 x}{x^{2} + 1} d x$ $3intx/(x^2+1)dx=3int1/2(2x)/(x^2+1)dx$

$= \frac{3}{2} \int \frac{1}{x^{2} + 1} 2 x d x = \frac{3}{2} \int \frac{1}{u} d u$ $=3/2int1/(x^2+1)2xdx=3/2int1/udu$

You can probably see how to finish to get $\frac{3}{2} ln (x^{2} + 1) + C$ $3/2ln(x^2+1)+C$ .

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How do I evaluate $\int \frac{3 x}{x^{2} + 1} d x$ $int(3x)/(x^2+1)dx$ ?

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How do I evaluate $\int \frac{3 x}{x^{2} + 1} d x$ $int(3x)/(x^2+1)dx$ ?