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

The integral of ((t-1)^2)/((2t+1)^2) is 1/4 t+(-12tln(1/2 |2t+1|)-6ln(1/2 |2t+1|)-9)/(8(2t+1))+C

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

integral

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

\frac{1}{4} t + \frac{- 12 t ln ( \frac{1}{2} ∣ 2 t + 1 ∣ ) - 6 ln ( \frac{1}{2} ∣ 2 t + 1 ∣ ) - 9}{8 ( 2 t + 1 )} + C

Solution steps

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

Long division

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

= \int \frac{1}{4} + \frac{- 12 t + 3}{4 ( 4 t ^{2} + 4 t + 1 )} d t

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{1}{4} d t + \int \frac{- 12 t + 3}{4 ( 4 t ^{2} + 4 t + 1 )} d t

\int \frac{1}{4} d t = \frac{1}{4} t

\int \frac{- 12 t + 3}{4 ( 4 t ^{2} + 4 t + 1 )} d t = \frac{- 12 t ln ( \frac{1}{2} ∣ 2 t + 1 ∣ ) - 6 ln ( \frac{1}{2} ∣ 2 t + 1 ∣ ) - 9}{8 ( 2 t + 1 )}

= \frac{1}{4} t + \frac{- 12 t ln ( \frac{1}{2} ∣ 2 t + 1 ∣ ) - 6 ln ( \frac{1}{2} ∣ 2 t + 1 ∣ ) - 9}{8 ( 2 t + 1 )}

Add a constant to the solution

= \frac{1}{4} t + \frac{- 12 t ln ( \frac{1}{2} ∣ 2 t + 1 ∣ ) - 6 ln ( \frac{1}{2} ∣ 2 t + 1 ∣ ) - 9}{8 ( 2 t + 1 )} + C

\frac{p}{( p ^{2} + 4 )}

4 x^{2} + 3 x + 2 \cdot 2 (x + 1)

(8 x - x^{4})^{3} (8 x^{3} - 16)

cos^{2} (\frac{x}{4})

(\frac{1}{x ^{4}}) sin (x)

What is the integral of ((t-1)^2)/((2t+1)^2) ?
The integral of ((t-1)^2)/((2t+1)^2) is 1/4 t+(-12tln(1/2 |2t+1|)-6ln(1/2 |2t+1|)-9)/(8(2t+1))+C