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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (t^3)/((t+1)^2)

integral

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

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

Solution steps

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

Apply u-substitution

: \int \frac{( u - 1 ) ^{3}}{u ^{2}} d u

= \int \frac{( u - 1 ) ^{3}}{u ^{2}} d u

Expand

\frac{( u - 1 ) ^{3}}{u ^{2}} : u - 3 + \frac{3}{u} - \frac{1}{u ^{2}}

= \int u - 3 + \frac{3}{u} - \frac{1}{u ^{2}} 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 u d u - \int 3 d u + \int \frac{3}{u} d u - \int \frac{1}{u ^{2}} d u

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

\int 3 d u = 3 u

\int \frac{3}{u} d u = 3 ln ∣ u ∣

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

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

Substitute back

u = t + 1

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

Simplify

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

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

Add a constant to the solution

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

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

x^{2} \cdot (\frac{1 - 2}{x})^{2}

(1 + x tan (x) - sin (x)) sec^{2} (x)

cos^{2} (x) - sin^{2} (x)

\frac{( 1 - x )}{( x + 2 )}

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