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

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

\int \frac{t ^{3}}{1 + t} d t

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

Solution steps

\int \frac{t ^{3}}{1 + t} d t

Apply u-substitution

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

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

Expand

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

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

\int u^{2} d u = \frac{u ^{3}}{3}

\int 3 u d u = \frac{3 u ^{2}}{2}

\int 3 d u = 3 u

\int \frac{1}{u} d u = ln ∣ u ∣

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

Substitute back

u = 1 + t

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

Add a constant to the solution

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

\int cos^{4} (x) sin^{5} (x) d x

2 x y^{^{'}} + y = 6 x, y (4) = 23

\frac{d}{d x} (6 x^{5} ln (x) + x^{5})

\frac{d y}{d x} = 8 + 8 y + x + x y

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

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