Solve ∫ (from 1 to 2) of 3/t^2-1dt

$$\int_{ 1 }^{ 2 } \frac{ 3 }{ { t }^{ 2 } } -1 d t$$

Evaluate

$$\frac{1}{2}=0.5$$

Solution Steps

Evaluate the indefinite integral first.

$$\int \frac{3}{t^{2}}-1\mathrm{d}t$$

Integrate the sum term by term.

$$\int \frac{3}{t^{2}}\mathrm{d}t+\int -1\mathrm{d}t$$

Factor out the constant in each of the terms.

$$3\int \frac{1}{t^{2}}\mathrm{d}t+\int -1\mathrm{d}t$$

Since \int t^{k}\mathrm{d}t=\frac{t^{k+1}}{k+1} for k\neq -1, replace \int \frac{1}{t^{2}}\mathrm{d}t with -\frac{1}{t}. Multiply 3 times -\frac{1}{t}.

$$-\frac{3}{t}+\int -1\mathrm{d}t$$

Find the integral of -1 using the table of common integrals rule \int a\mathrm{d}t=at.

$$-\frac{3}{t}-t$$

The definite integral is the antiderivative of the expression evaluated at the upper limit of integration minus the antiderivative evaluated at the lower limit of integration.

$$-3\times 2^{-1}-2-\left(-3\times 1^{-1}-1\right)$$

Simplify.

$$\frac{1}{2}$$