Solve ∫ (from 0 to 1) of 1-t^2/1+6t^2+t^4dt=1/2I

$$\int _ { 0 } ^ { 1 } \frac { 1 - t ^ { 2 } } { 1 + 6 t ^ { 2 } + t ^ { 4 } } d t = \frac { 1 } { 2 } I$$

Solve for I

$$I=2\int _{0}^{1}\frac{1-t^{2}}{t^{4}+6t^{2}+1}\mathrm{d}t$$

Steps for Solving Linear Equation

Swap sides so that all variable terms are on the left hand side.

$$\frac{1}{2}I=\int _{0}^{1}\frac{1-t^{2}}{1+6t^{2}+t^{4}}\mathrm{d}t$$

The equation is in standard form.

$$\frac{1}{2}I=\int _{0}^{1}\frac{1-t^{2}}{t^{4}+6t^{2}+1}\mathrm{d}t$$

Multiply both sides by 2.

$$\frac{\frac{1}{2}I}{\frac{1}{2}}=\frac{\int _{0}^{1}\frac{1-t^{2}}{t^{4}+6t^{2}+1}\mathrm{d}t}{\frac{1}{2}}$$

Dividing by \frac{1}{2} undoes the multiplication by \frac{1}{2}.

$$I=\frac{\int _{0}^{1}\frac{1-t^{2}}{t^{4}+6t^{2}+1}\mathrm{d}t}{\frac{1}{2}}$$

Divide \int _{0}^{1}\frac{1-t^{2}}{1+6t^{2}+t^{4}}\mathrm{d}t by \frac{1}{2} by multiplying \int _{0}^{1}\frac{1-t^{2}}{1+6t^{2}+t^{4}}\mathrm{d}t by the reciprocal of \frac{1}{2}.

$$I=2\int _{0}^{1}\frac{1-t^{2}}{t^{4}+6t^{2}+1}\mathrm{d}t$$