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

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

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integral of t/(sqrt(t^2+2t-3))

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

t^{2} + 2 t - 3 - ln (\frac{1}{2} t^{2} + 2 t - 3 + t + 1) + C

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

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

Expand

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

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

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

= u^{2} - 4 - ln (\frac{1}{2} u^{2} - 4 + u)

Substitute back

u = t + 1

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

Simplify

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

= t^{2} + 2 t - 3 - ln (\frac{1}{2} t^{2} + 2 t - 3 + t + 1)

Add a constant to the solution

= t^{2} + 2 t - 3 - ln (\frac{1}{2} t^{2} + 2 t - 3 + t + 1) + C

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