Solve J ( x + 3 ) ^ 2 = ∫ 144

$$J ( x + 3 ) ^ { 2 } = \int 144$$

Solve for J

$$\left\{\begin{matrix}J=\frac{144x+С}{\left(x+3\right)^{2}}\text{, }&x\neq -3\\J\in \mathrm{R}\text{, }&x=-3\text{ and }С=432\end{matrix}\right.$$

Steps for Solving Linear Equation

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

$$J\left(x^{2}+6x+9\right)=\int 144\mathrm{d}x$$

Use the distributive property to multiply J by x^{2}+6x+9.

$$Jx^{2}+6Jx+9J=\int 144\mathrm{d}x$$

Combine all terms containing J.

$$\left(x^{2}+6x+9\right)J=\int 144\mathrm{d}x$$

The equation is in standard form.

$$\left(x^{2}+6x+9\right)J=144x+С$$

Divide both sides by x^{2}+6x+9.

$$\frac{\left(x^{2}+6x+9\right)J}{x^{2}+6x+9}=\frac{144x+С}{x^{2}+6x+9}$$

Dividing by x^{2}+6x+9 undoes the multiplication by x^{2}+6x+9.

$$J=\frac{144x+С}{x^{2}+6x+9}$$

Divide 144x+С by x^{2}+6x+9.

$$J=\frac{144x+С}{\left(x+3\right)^{2}}$$