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

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

Evaluate

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

Short Solution Steps

Factor 1-a^{2}.

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

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{a}{\left(a-1\right)\left(-a-1\right)} times \frac{a^{2}+1}{a^{2}+1}. Multiply \frac{a}{1+a^{2}} times \frac{\left(a-1\right)\left(-a-1\right)}{\left(a-1\right)\left(-a-1\right)}.

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

Since \frac{a\left(a^{2}+1\right)}{\left(a-1\right)\left(-a-1\right)\left(a^{2}+1\right)} and \frac{a\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.

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

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

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

Combine like terms in a^{3}+a-a^{3}-a^{2}+a^{2}+a.

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

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

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

Factor

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