ye1neh4

2022-09-27

Use a table of integrals to evaluate the following integrals. $$\int \frac{x}{x+1}dx$$

vyhlodatis

Beginner2022-09-28Added 14 answers

$$\int \frac{x}{x+1}dx$$

Formula: $$\int \frac{udu}{a+bu}=\frac{1}{{b}^{2}}(a+bu-a\mathrm{ln}|a+bu|)+C$$

$$u=x\Rightarrow du=x$$, and a=1, b=1

Therefore,

$$\int \frac{x}{x+1}dx=\underset{\int \frac{udu}{a+bu}}{\underset{\u23df}{\int \frac{x}{1+1x}dx}}=\frac{1}{(1{)}^{2}}(1+1x-1\mathrm{ln}|1+1x|)+C$$

Simplify

$$\int \frac{x}{x+1}dx=1+x-\mathrm{ln}|1+x|+C$$

$$\int \frac{x}{x+1}dx=x-\mathrm{ln}|x+1|+C$$

Result:

$$x-\mathrm{ln}|x+1|+C$$

Formula: $$\int \frac{udu}{a+bu}=\frac{1}{{b}^{2}}(a+bu-a\mathrm{ln}|a+bu|)+C$$

$$u=x\Rightarrow du=x$$, and a=1, b=1

Therefore,

$$\int \frac{x}{x+1}dx=\underset{\int \frac{udu}{a+bu}}{\underset{\u23df}{\int \frac{x}{1+1x}dx}}=\frac{1}{(1{)}^{2}}(1+1x-1\mathrm{ln}|1+1x|)+C$$

Simplify

$$\int \frac{x}{x+1}dx=1+x-\mathrm{ln}|1+x|+C$$

$$\int \frac{x}{x+1}dx=x-\mathrm{ln}|x+1|+C$$

Result:

$$x-\mathrm{ln}|x+1|+C$$

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