What is the integral of (sqrt(t^2+2t-3))/((t+1)^4) ?

The integral of (sqrt(t^2+2t-3))/((t+1)^4) is ((t^2+2t-3)^{3/2})/(12(t+1)^3)+C

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integral of (sqrt(t^2+2t-3))/((t+1)^4)

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

\frac{( t ^{2} + 2 t - 3 ) ^{\frac{3}{2}}}{12 ( t + 1 ) ^{3}} + C

Solution steps

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

Apply u-substitution

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

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

Apply Trigonometric Substitution

: \int \frac{tan ^{2} ( v )}{4 sec ^{3} ( v )} d v

= \int \frac{tan ^{2} ( v )}{4 sec ^{3} ( v )} d v

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{tan ^{2} ( v )}{sec ^{3} ( v )} d v

Rewrite using trig identities

= \frac{1}{4} \cdot \int sin^{2} (v) cos (v) d v

Apply u-substitution

: \int w^{2} d w

= \frac{1}{4} \cdot \int w^{2} d w

Apply the Power Rule

: \frac{w ^{3}}{3}

= \frac{1}{4} \cdot \frac{w ^{3}}{3}

Substitute back

= \frac{1}{4} \cdot \frac{sin ^{3} ( arcsec ( \frac{1}{2} ( t + 1 ) ) )}{3}

Simplify

\frac{1}{4} \cdot \frac{sin ^{3} ( arcsec ( \frac{1}{2} ( t + 1 ) ) )}{3} : \frac{( t ^{2} + 2 t - 3 ) ^{\frac{3}{2}}}{12 ( t + 1 ) ^{3}}

= \frac{( t ^{2} + 2 t - 3 ) ^{\frac{3}{2}}}{12 ( t + 1 ) ^{3}}

Add a constant to the solution

= \frac{( t ^{2} + 2 t - 3 ) ^{\frac{3}{2}}}{12 ( t + 1 ) ^{3}} + C

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What is the integral of (sqrt(t^2+2t-3))/((t+1)^4) ?
The integral of (sqrt(t^2+2t-3))/((t+1)^4) is ((t^2+2t-3)^{3/2})/(12(t+1)^3)+C