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

The integral of t(1-ln^2(t)) is (t^2)/2-1/2 t^2ln^2(t)+1/2 t^2ln(t)-(t^2)/4+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of t(1-ln^2(t))

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

\frac{t ^{2}}{2} - \frac{1}{2} t^{2} ln^{2} (t) + \frac{1}{2} t^{2} ln (t) - \frac{t ^{2}}{4} + C

Solution steps

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

Expand

t (1 - ln^{2} (t)) : t - t ln^{2} (t)

= \int t - t ln^{2} (t) 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 t d t - \int t ln^{2} (t) d t

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

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

= \frac{t ^{2}}{2} - (\frac{1}{2} t^{2} ln^{2} (t) - \frac{1}{2} t^{2} ln (t) + \frac{t ^{2}}{4})

Simplify

= \frac{t ^{2}}{2} - \frac{1}{2} t^{2} ln^{2} (t) + \frac{1}{2} t^{2} ln (t) - \frac{t ^{2}}{4}

Add a constant to the solution

= \frac{t ^{2}}{2} - \frac{1}{2} t^{2} ln^{2} (t) + \frac{1}{2} t^{2} ln (t) - \frac{t ^{2}}{4} + C

\int (- 2 sin (x) + 4 cos (x)) d x

\int x e^{- (x + y)} d x

\frac{d}{d x} ((7 - 3 x) (x + 5))

y^{^{'}} + 3 y = cos (x)

4 y^{^{''}} + y^{^{'}} + 8 y = 0

What is the integral of t(1-ln^2(t)) ?
The integral of t(1-ln^2(t)) is (t^2)/2-1/2 t^2ln^2(t)+1/2 t^2ln(t)-(t^2)/4+C