# Evaluate the integral. \int \frac{x+3}{x-1}dx

Evaluate the integral.
$$\displaystyle\int{\frac{{{x}+{3}}}{{{x}-{1}}}}{\left.{d}{x}\right.}$$

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Lible1953
Step 1
Given: $$\displaystyle{I}=\int{\frac{{{x}+{3}}}{{{x}-{1}}}}{\left.{d}{x}\right.}$$
For evaluating given integral, first we simplify given expression then integrate it
Step 2
So,
$$\displaystyle{I}=\int{\frac{{{x}+{3}}}{{{x}-{1}}}}{\left.{d}{x}\right.}$$
$$\displaystyle=\int{\frac{{{\left({x}-{1}+{4}\right)}}}{{{\left({x}-{1}\right)}}}}{\left.{d}{x}\right.}$$
$$\displaystyle=\int{\left({\frac{{{x}-{1}}}{{{x}-{1}}}}+{\frac{{{4}}}{{{x}-{1}}}}\right)}{\left.{d}{x}\right.}$$
$$\displaystyle=\int{\left({1}+{\frac{{{4}}}{{{x}-{1}}}}\right)}{\left.{d}{x}\right.}$$
$$\displaystyle=\int{\left.{d}{x}\right.}+{4}\int{\frac{{{\left.{d}{x}\right.}}}{{{x}-{1}}}}$$
$$\displaystyle{\left(\because\int{\left.{d}{x}\right.}={x}+{c},\int{\frac{{{\left.{d}{x}\right.}}}{{{x}-{a}}}}={\ln}{\left|{x}-{a}\right|}+{c}\right)}$$
$$\displaystyle={x}+{4}{\ln}{\left|{x}-{1}\right|}+{c}$$
Hence, given integral is equal to $$\displaystyle{\left({x}+{4}{\ln}{\left|{x}-{1}\right|}+{c}\right)}$$.
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Charles Clute
Step 1
Use Integration by Substitution.
Let u=x-1, du=dx, then x+3dx=u+1+3 du
Step 2
Using u and du above, rewrite $$\displaystyle\int{\frac{{{x}+{3}}}{{{x}-{1}}}}{\left.{d}{x}\right.}$$.
$$\displaystyle\int{\left({u}+{1}+{3}\right)}\times{\frac{{{1}}}{{{u}}}}{d}{u}$$
Step 3
Use Sum Rule: $$\displaystyle\int{f{{\left({x}\right)}}}+{g{{\left({x}\right)}}}{\left.{d}{x}\right.}=\int{f{{\left({x}\right)}}}{\left.{d}{x}\right.}+\int{g{{\left({x}\right)}}}{\left.{d}{x}\right.}$$.
$$\displaystyle\int{1}{d}{u}+\int{\frac{{{4}}}{{{u}}}}{d}{u}$$
Step 4
Use this rule: $$\displaystyle\int{a}{\left.{d}{x}\right.}={a}{x}+{C}$$.
$$\displaystyle{u}+\int{\frac{{{4}}}{{{u}}}}{d}{u}$$
Step 5
Use Constant Factor Rule: $$\displaystyle\int{c}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={c}\int{f{{\left({x}\right)}}}{\left.{d}{x}\right.}$$.
$$\displaystyle{u}+{4}\int{\frac{{{1}}}{{{u}}}}{d}{u}$$
Step 6
The derivative of $$\displaystyle{\ln{{x}}}\ {i}{s}\ {\frac{{{1}}}{{{x}}}}$$.
$$\displaystyle{u}+{4}{\ln{{u}}}$$
Step 7
Substitute u=x−1 back into the original integral.
$$\displaystyle{x}-{1}+{4}{\ln{{\left({x}-{1}\right)}}}$$
Step 8
$$\displaystyle{x}-{1}+{4}{\ln{{\left({x}-{1}\right)}}}+{C}$$
Step 9
Merge numbers into the constant.
$$\displaystyle{x}+{4}{\ln{{\left({x}-{1}\right)}}}+{C}$$