What is the integral of 1/(2at+t^2) ?

The integral of 1/(2at+t^2) is -1/(2a)(ln|(t+a)/a+1|-ln|(t+a)/a-1|)+C

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integral of 1/(2at+t^2)

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

- \frac{1}{2 a} (ln \frac{t + a}{a} + 1 - ln \frac{t + a}{a} - 1) + C

Solution steps

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

Complete the square

2 a t + t^{2} : (t + a)^{2} - a^{2}

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

Apply u-substitution

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

= \int \frac{1}{u ^{2} - a ^{2}} d u

Apply Integral Substitution

= \int \frac{1}{a ( v ^{2} - 1 )} d v

Take the constant out:

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

= \frac{1}{a} \cdot \int \frac{1}{v ^{2} - 1} d v

Factor

v^{2} - 1 : - (- v^{2} + 1)

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

Take the constant out:

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

= \frac{1}{a} (- \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}

= \frac{1}{a} (- (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}))

Substitute back

= \frac{1}{a} (- (\frac{ln \frac{t + a}{a} + 1}{2} - \frac{ln \frac{t + a}{a} - 1}{2}))

Simplify

\frac{1}{a} (- (\frac{ln \frac{t + a}{a} + 1}{2} - \frac{ln \frac{t + a}{a} - 1}{2})) : - \frac{1}{2 a} (ln \frac{t + a}{a} + 1 - ln \frac{t + a}{a} - 1)

= - \frac{1}{2 a} (ln \frac{t + a}{a} + 1 - ln \frac{t + a}{a} - 1)

Add a constant to the solution

= - \frac{1}{2 a} (ln \frac{t + a}{a} + 1 - ln \frac{t + a}{a} - 1) + C

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What is the integral of 1/(2at+t^2) ?
The integral of 1/(2at+t^2) is -1/(2a)(ln|(t+a)/a+1|-ln|(t+a)/a-1|)+C