integral of 1/(2-t)+1/(2+t) — step-by-step solution

Let's compute the integral:

\int (\frac{1}{2 - t} + \frac{1}{2 + t}) d t

ⅰ

\int (\frac{1}{2 - t} + \frac{1}{2 + t}) d t

\int \frac{1}{2 - t} d t + \int \frac{1}{2 + t} d t

\int \frac{1}{2 - t} d t + \int \frac{1}{2 + t} d t

Let

u = 2 - t

. Then

du = - dt

, so

dt = - du

\int \frac{1}{2 - t} d t = -1 \cdot \int \frac{1}{u} d u

We use

\int \frac{1}{u} d u = \ln | u |

. The absolute value covers both signs of the denominator, on any interval where

2 - t \neq 0

Substitute

u

2 - t

back into the antiderivative

Let

u = 2 + t

. Then

du = dt

, so

dt = du

\int \frac{1}{2 + t} d t = \int \frac{1}{u} d u

We use

\int \frac{1}{u} d u = \ln | u |

. The absolute value covers both signs of the denominator, on any interval where

2 + t \neq 0

Substitute

u

2 + t

back into the antiderivative

ⓘ The derivative of

\ln | 2 + t |

\frac{1}{2 + t}

, so

\ln | 2 + t |

is an antiderivative of

\frac{1}{2 + t}

\int \frac{1}{2 + t} d t

\ln | 2 + t |

- 1 \cdot \ln | 2 - t | + \ln | 2 + t |

ⓘ Multiplying by 1 changes nothing:

1 \cdot \ln | 2 - t |

\ln | 2 - t |

- \ln | 2 - t | + \ln | 2 + t |

ⓘ For strictly positive arguments, a sum of logarithms is the logarithm of the product, a difference that of the quotient, ln(a) + ln(b) = ln(a*b):

- \ln | 2 - t | + \ln | 2 + t |

\ln (\frac{| 2 + t |}{| 2 - t |})

\ln (\frac{| 2 + t |}{| 2 - t |})

ⓘ The quotient of absolute values is the absolute value of the quotient:

\frac{| 2 + t |}{| 2 - t |}

| \frac{2 + t}{2 - t} |

\ln | \frac{2 + t}{2 - t} |

▶Indefinite integrals are defined up to an additive constant C, so this yields:

\ln | \frac{t + 2}{2 - t} | + C