Solve ∫ (from 1 to 2) of x^2+1/x^4 wrt x=21/a

$$\int _ { 1 } ^ { 2 } x ^ { 2 } + \frac { 1 } { x ^ { 4 } } d x = \frac { 21 } { a }$$

Solve for a

$$a=8$$

$$x\neq 0$$

Steps for Solving Linear Equation

Variable a cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by a.

$$a\int _{1}^{2}x^{2}+\frac{1}{x^{4}}\mathrm{d}x=21$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x^{2} times \frac{x^{4}}{x^{4}}.

$$a\int _{1}^{2}\frac{x^{2}x^{4}}{x^{4}}+\frac{1}{x^{4}}\mathrm{d}x=21$$

Since \frac{x^{2}x^{4}}{x^{4}} and \frac{1}{x^{4}} have the same denominator, add them by adding their numerators.

$$a\int _{1}^{2}\frac{x^{2}x^{4}+1}{x^{4}}\mathrm{d}x=21$$

Do the multiplications in x^{2}x^{4}+1.

$$a\int _{1}^{2}\frac{x^{6}+1}{x^{4}}\mathrm{d}x=21$$

The equation is in standard form.

$$\frac{21}{8}a=21$$

Divide both sides of the equation by \frac{21}{8}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{21}{8}a}{\frac{21}{8}}=\frac{21}{\frac{21}{8}}$$

Dividing by \frac{21}{8} undoes the multiplication by \frac{21}{8}.

$$a=\frac{21}{\frac{21}{8}}$$

Divide 21 by \frac{21}{8} by multiplying 21 by the reciprocal of \frac{21}{8}.

$$a=8$$