What is the integral of (e^{2t}+1/((t+1)^2)) ?

The integral of (e^{2t}+1/((t+1)^2)) is 1/2 e^{2t}-1/(t+1)+C

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integral of (e^{2t}+1/((t+1)^2))

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

\frac{1}{2} e^{2 t} - \frac{1}{t + 1} + C

Solution steps

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

\int e^{2 t} d t = \frac{1}{2} e^{2 t}

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

= \frac{1}{2} e^{2 t} - \frac{1}{t + 1}

Add a constant to the solution

= \frac{1}{2} e^{2 t} - \frac{1}{t + 1} + C

\frac{s - 3}{s ^{2} + 10 s + 9}

x \to 4 lim (2 x + 1)

\int \frac{1}{- u ^{2} + 1} d u

\frac{( x ^{8} )}{8 ( 1 - x ^{2} ) ^{4}}

\frac{d}{d x} (\frac{1}{( 9 + x ) ^{2}})

What is the integral of (e^{2t}+1/((t+1)^2)) ?
The integral of (e^{2t}+1/((t+1)^2)) is 1/2 e^{2t}-1/(t+1)+C