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

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

Evaluate

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

Short Solution Steps

Factor 1-a^{2}.

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

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

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

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

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

Do the multiplications in a^{2}+1+\left(a-1\right)\left(-a-1\right).

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

Combine like terms in a^{2}+1-a^{2}-a+a+1.

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

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

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

Differentiate w.r.t. a

$$\frac{8a^{3}}{\left(1-a^{4}\right)^{2}}$$