What is the integral from 0 to 6x^2 of (t/(t+36)) ?

The integral from 0 to 6x^2 of (t/(t+36)) is 6x^2-36(ln|6x^2+36|-2ln(6))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from 0 to 6x^2 of (t/(t+36))

\int_{0}^{6 x^{2}} (\frac{t}{t + 36}) d t

6 x^{2} - 36 (ln 6 x^{2} + 36 - 2 ln (6))

Solution steps

\int_{0}^{6 x^{2}} \frac{t}{t + 36} d t

Apply u-substitution

= \int_{36}^{6 x^{2} + 36} \frac{u - 36}{u} d u

Expand

\frac{u - 36}{u} : 1 - \frac{36}{u}

= \int_{36}^{6 x^{2} + 36} 1 - \frac{36}{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_{36}^{6 x^{2} + 36} 1 d u - \int_{36}^{6 x^{2} + 36} \frac{36}{u} d u

\int_{36}^{6 x^{2} + 36} 1 d u = 6 x^{2}

\int_{36}^{6 x^{2} + 36} \frac{36}{u} d u = 36 (ln 6 x^{2} + 36 - 2 ln (6))

= 6 x^{2} - 36 (ln 6 x^{2} + 36 - 2 ln (6))

\frac{d}{d x} (20 - x)

\int x^{\frac{6}{7}} d x

d er i v a t i v e 5 x (sin (x) + cos (x))

\frac{d}{d θ} (2 sin (2 θ))

t an g e n t f (x) = 6 + 5 x^{2} - 2 x^{3}, at x = 1

What is the integral from 0 to 6x^2 of (t/(t+36)) ?
The integral from 0 to 6x^2 of (t/(t+36)) is 6x^2-36(ln|6x^2+36|-2ln(6))