What is the integral of (200)/(t^2+3t+2) ?

The integral of (200)/(t^2+3t+2) is -400(1/2 ln|2t+4|-1/2 ln|2t+2|)+C

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integral of (200)/(t^2+3t+2)

\int \frac{200}{t ^{2} + 3 t + 2} d t

- 400 (\frac{1}{2} ln ∣ 2 t + 4 ∣ - \frac{1}{2} ln ∣ 2 t + 2 ∣) + C

Solution steps

\int \frac{200}{t ^{2} + 3 t + 2} d t

Take the constant out:

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

= 200 \cdot \int \frac{1}{t ^{2} + 3 t + 2} d t

Complete the square

t^{2} + 3 t + 2 : (t + \frac{3}{2})^{2} - \frac{1}{4}

= 200 \cdot \int \frac{1}{( t + \frac{3}{2} ) ^{2} - \frac{1}{4}} d t

Apply u-substitution

: \int \frac{4}{4 u ^{2} - 1} d u

= 200 \cdot \int \frac{4}{4 u ^{2} - 1} d u

Take the constant out:

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

= 200 \cdot 4 \cdot \int \frac{1}{4 u ^{2} - 1} d u

Factor

4 u^{2} - 1 : - (- 4 u^{2} + 1)

= 200 \cdot 4 \cdot \int \frac{1}{- ( - 4 u ^{2} + 1 )} d u

Take the constant out:

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

= 200 \cdot 4 (- \int \frac{1}{- 4 u ^{2} + 1} d u)

Apply Integral Substitution

= 200 \cdot 4 (- \int \frac{1}{2 ( - v ^{2} + 1 )} d v)

Take the constant out:

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

= 200 \cdot 4 (- \frac{1}{2} \cdot \int \frac{1}{- v ^{2} + 1} d v)

Use the common integral:

\int \frac{1}{- v ^{2} + 1} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

= 200 \cdot 4 (- \frac{1}{2} (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}))

Substitute back

= 200 \cdot 4 (- \frac{1}{2} (\frac{ln 2 ( t + \frac{3}{2} ) + 1}{2} - \frac{ln 2 ( t + \frac{3}{2} ) - 1}{2}))

Simplify

200 \cdot 4 (- \frac{1}{2} (\frac{ln 2 ( t + \frac{3}{2} ) + 1}{2} - \frac{ln 2 ( t + \frac{3}{2} ) - 1}{2})) : - 400 (\frac{1}{2} ln ∣ 2 t + 4 ∣ - \frac{1}{2} ln ∣ 2 t + 2 ∣)

= - 400 (\frac{1}{2} ln ∣ 2 t + 4 ∣ - \frac{1}{2} ln ∣ 2 t + 2 ∣)

Add a constant to the solution

= - 400 (\frac{1}{2} ln ∣ 2 t + 4 ∣ - \frac{1}{2} ln ∣ 2 t + 2 ∣) + C

\frac{d y}{d x} = x x - 12

\int \frac{3 x ^{3}}{5 - 9 x ^{4}} d x

\int \frac{x ^{2} + 6 x + 2}{x ^{3} + 9 x ^{2} + 6 x + 3} d x

- 3 cos (x), at x = - \frac{π}{3}

e^{2 x^{3}}

What is the integral of (200)/(t^2+3t+2) ?
The integral of (200)/(t^2+3t+2) is -400(1/2 ln|2t+4|-1/2 ln|2t+2|)+C