Solve ∫ 3/(1+x)^5 wrt x=m(1+x)^n+c

$$\int \frac { 3 } { ( 1 + x ) ^ { 5 } } d x = m ( 1 + x ) ^ { n } + c$$

Solve for m

$$m=\frac{\left(-4c\left(x+1\right)^{4}-3\right)\left(x+1\right)^{-n-4}}{4}+\frac{С}{\left(x+1\right)^{n}}$$

$$x>-1\text{ or }\left(Denominator(n)\text{bmod}2=1\text{ and }x<-1\right)$$

Steps for Solving Linear Equation

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

$$m\left(1+x\right)^{n}+c=\int \frac{3}{\left(1+x\right)^{5}}\mathrm{d}x$$

Subtract c from both sides.

$$m\left(1+x\right)^{n}=\int \frac{3}{\left(1+x\right)^{5}}\mathrm{d}x-c$$

The equation is in standard form.

$$\left(x+1\right)^{n}m=-c-\frac{3}{4\left(x+1\right)^{4}}+С$$

Divide both sides by \left(1+x\right)^{n}.

$$\frac{\left(x+1\right)^{n}m}{\left(x+1\right)^{n}}=\frac{-c-\frac{3}{4\left(x+1\right)^{4}}+С}{\left(x+1\right)^{n}}$$

Dividing by \left(1+x\right)^{n} undoes the multiplication by \left(1+x\right)^{n}.

$$m=\frac{-c-\frac{3}{4\left(x+1\right)^{4}}+С}{\left(x+1\right)^{n}}$$

Divide -c+С-\frac{3}{4\left(1+x\right)^{4}} by \left(1+x\right)^{n}.

$$m=\frac{С}{\left(x+1\right)^{n}}-\frac{c}{\left(x+1\right)^{n}}-\frac{3\left(x+1\right)^{-n-4}}{4}$$

Solve for c

$$c=-m\left(x+1\right)^{n}-\frac{3}{4\left(x+1\right)^{4}}+С$$

$$x>-1\text{ or }\left(Denominator(n)\text{bmod}2=1\text{ and }x<-1\right)$$