Solve 1/1-2a+1/1+2a-1/4a^2-1

$$\frac { 1 } { 1 - 2 a } + \frac { 1 } { 1 + 2 a } - \frac { 1 } { 4 a ^ { 2 } - 1 }$$

Evaluate

$$\frac{3}{1-4a^{2}}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 1-2a and 1+2a is \left(-2a+1\right)\left(2a+1\right). Multiply \frac{1}{1-2a} times \frac{2a+1}{2a+1}. Multiply \frac{1}{1+2a} times \frac{-2a+1}{-2a+1}.

$$\frac{2a+1}{\left(-2a+1\right)\left(2a+1\right)}+\frac{-2a+1}{\left(-2a+1\right)\left(2a+1\right)}-\frac{1}{4a^{2}-1}$$

Since \frac{2a+1}{\left(-2a+1\right)\left(2a+1\right)} and \frac{-2a+1}{\left(-2a+1\right)\left(2a+1\right)} have the same denominator, add them by adding their numerators.

$$\frac{2a+1-2a+1}{\left(-2a+1\right)\left(2a+1\right)}-\frac{1}{4a^{2}-1}$$

Combine like terms in 2a+1-2a+1.

$$\frac{2}{\left(-2a+1\right)\left(2a+1\right)}-\frac{1}{4a^{2}-1}$$

Factor 4a^{2}-1.

$$\frac{2}{\left(-2a+1\right)\left(2a+1\right)}-\frac{1}{\left(2a-1\right)\left(2a+1\right)}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(-2a+1\right)\left(2a+1\right) and \left(2a-1\right)\left(2a+1\right) is \left(2a-1\right)\left(2a+1\right). Multiply \frac{2}{\left(-2a+1\right)\left(2a+1\right)} times \frac{-1}{-1}.

$$\frac{2\left(-1\right)}{\left(2a-1\right)\left(2a+1\right)}-\frac{1}{\left(2a-1\right)\left(2a+1\right)}$$

Since \frac{2\left(-1\right)}{\left(2a-1\right)\left(2a+1\right)} and \frac{1}{\left(2a-1\right)\left(2a+1\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{2\left(-1\right)-1}{\left(2a-1\right)\left(2a+1\right)}$$

Do the multiplications in 2\left(-1\right)-1.

$$\frac{-2-1}{\left(2a-1\right)\left(2a+1\right)}$$

Do the calculations in -2-1.

$$\frac{-3}{\left(2a-1\right)\left(2a+1\right)}$$

Expand \left(2a-1\right)\left(2a+1\right).

$$\frac{-3}{4a^{2}-1}$$

Differentiate w.r.t. a

$$\frac{24a}{\left(1-4a^{2}\right)^{2}}$$