Math / Calculus

Question $\int_{0}^{1} \frac{3 x}{4 x^{2}-1} d x$

Studdy Solution

STEP 1

1. The integral $\int_{0}^{1} \frac{3x}{4x^{2}-1} \, dx$ is a rational function.
2. The integrand can be simplified using substitution.
3. The limits of integration will be adjusted according to the substitution used.

STEP 2

1. Choose an appropriate substitution to simplify the integrand.
2. Adjust the limits of integration based on the substitution.
3. Integrate the simplified function.
4. Substitute back to the original variable, if necessary, and evaluate the definite integral.

STEP 3

Choose a substitution that simplifies the denominator. Let $u = 4x^2 - 1$ . Then, compute $du$ :
$u = 4x^2 - 1$ $\frac{du}{dx} = 8x$ $du = 8x \, dx$
To express $3x \, dx$ in terms of $du$ , we have:
$3x \, dx = \frac{3}{8} du$

STEP 4

Adjust the limits of integration according to the substitution. When $x = 0$ , $u = 4(0)^2 - 1 = -1$ . When $x = 1$ , $u = 4(1)^2 - 1 = 3$ .
Thus, the new limits of integration are from $u = -1$ to $u = 3$ .

STEP 5

Substitute into the integral:
$\int_{0}^{1} \frac{3x}{4x^2 - 1} \, dx = \int_{-1}^{3} \frac{3}{8} \frac{1}{u} \, du$
This simplifies to:
$\frac{3}{8} \int_{-1}^{3} \frac{1}{u} \, du$
The antiderivative of $\frac{1}{u}$ is $\ln|u|$ , so we have:
$\frac{3}{8} \left[ \ln|u| \right]_{-1}^{3}$

STEP 6

Evaluate the definite integral:
$\frac{3}{8} \left[ \ln|3| - \ln|-1| \right]$
Since $\ln|-1|$ is undefined in the real number system, we consider the principal value, which results in:
$\frac{3}{8} \left[ \ln 3 - \ln 1 \right]$
Since $\ln 1 = 0$ , this simplifies to:
$\frac{3}{8} \ln 3$
The value of the integral is:
$\boxed{\frac{3}{8} \ln 3}$

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