What is the integral from 0 to 4 of (2t)/(t^2-25) ?

The integral from 0 to 4 of (2t)/(t^2-25) is 2ln(3)-2ln(5)

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integral from 0 to 4 of (2t)/(t^2-25)

\int_{0}^{4} \frac{2 t}{t ^{2} - 25} d t

2 ln (3) - 2 ln (5)

Decimal Notation

- 1.02165 \dots

Solution steps

\int_{0}^{4} \frac{2 t}{t ^{2} - 25} d t

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int_{0}^{4} \frac{t}{t ^{2} - 25} d t

Apply u-substitution

: \int_{- 25}^{- 9} \frac{1}{2 u} d u

= 2 \cdot \int_{- 25}^{- 9} \frac{1}{2 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \frac{1}{2} \cdot \int_{- 25}^{- 9} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 \cdot \frac{1}{2} [ln ∣ u ∣]_{- 25}^{- 9}

Simplify

2 \cdot \frac{1}{2} [ln ∣ u ∣]_{- 25}^{- 9} : [ln ∣ u ∣]_{- 25}^{- 9}

= [ln ∣ u ∣]_{- 25}^{- 9}

Compute the boundaries:

2 ln (3) - 2 ln (5)

= 2 ln (3) - 2 ln (5)

\int_{\frac{π}{4}}^{\frac{π}{2}} 2 - cot^{2} (x) d x

\int_{0}^{1} 4 x^{3} x^{2} + 1 d x

\int_{1}^{2} \frac{x ^{2}}{2} + \frac{1}{2 x ^{2}} d x

\int_{- 1}^{1} \frac{5 x}{( x + 5 ) ^{2}} d x

\int_{0}^{r} (b r^{2} + cr) \cdot 2 π r d r

What is the integral from 0 to 4 of (2t)/(t^2-25) ?
The integral from 0 to 4 of (2t)/(t^2-25) is 2ln(3)-2ln(5)