What is the integral of (10t)/((t+1)^2) ?

The integral of (10t)/((t+1)^2) is 10(ln|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 (10t)/((t+1)^2)

integral

\frac{10 t}{( t + 1 ) ^{2}}

10 (ln ∣ t + 1 ∣ + \frac{1}{t + 1}) + C

Solution steps

\int \frac{10 t}{( t + 1 ) ^{2}} d t

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 10 \cdot \int \frac{t}{( t + 1 ) ^{2}} d t

Apply u-substitution

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

= 10 \cdot \int \frac{u - 1}{u ^{2}} d u

Expand

\frac{u - 1}{u ^{2}} : \frac{1}{u} - \frac{1}{u ^{2}}

= 10 \cdot \int \frac{1}{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

= 10 (\int \frac{1}{u} d u - \int \frac{1}{u ^{2}} d u)

\int \frac{1}{u} d u = ln ∣ u ∣

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

= 10 (ln ∣ u ∣ - (- \frac{1}{u}))

Substitute back

u = t + 1

= 10 (ln ∣ t + 1 ∣ - (- \frac{1}{t + 1}))

Simplify

= 10 (ln ∣ t + 1 ∣ + \frac{1}{t + 1})

Add a constant to the solution

= 10 (ln ∣ t + 1 ∣ + \frac{1}{t + 1}) + C

8 + x

\frac{( a ^{x} )}{( a ^{x} - 1 )}

sec^{4} (x) tan^{2} (x)

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

3 x^{2} + 6 x - 2

What is the integral of (10t)/((t+1)^2) ?
The integral of (10t)/((t+1)^2) is 10(ln|t+1|+1/(t+1))+C