What is the integral from 0 to 1 of (4x+6)/((x+1)^2) ?

The integral from 0 to 1 of (4x+6)/((x+1)^2) is 4ln(2)+1

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integral from 0 to 1 of (4x+6)/((x+1)^2)

\int_{0}^{1} \frac{4 x + 6}{( x + 1 ) ^{2}} d x

4 ln (2) + 1

Decimal Notation

3.77258 \dots

Solution steps

\int_{0}^{1} \frac{4 x + 6}{( x + 1 ) ^{2}} d x

Expand

\frac{4 x + 6}{( x + 1 ) ^{2}} : \frac{4 x}{( x + 1 ) ^{2}} + \frac{6}{( x + 1 ) ^{2}}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{1} \frac{4 x}{( x + 1 ) ^{2}} d x + \int_{0}^{1} \frac{6}{( x + 1 ) ^{2}} d x

\int_{0}^{1} \frac{4 x}{( x + 1 ) ^{2}} d x = 4 (ln (2) - \frac{1}{2})

\int_{0}^{1} \frac{6}{( x + 1 ) ^{2}} d x = 3

= 4 (ln (2) - \frac{1}{2}) + 3

Simplify

4 (ln (2) - \frac{1}{2}) + 3 : 4 ln (2) + 1

= 4 ln (2) + 1

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

\int \frac{3 x ^{2} - 2}{x ^{3} - 2 x - 8} d x

t \to 3 + lim (ln (t - 3))

x \to - \frac{π}{4} lim (csc (x))

n = 1 \sum \infty (\frac{1 + n}{n})^{n}

What is the integral from 0 to 1 of (4x+6)/((x+1)^2) ?
The integral from 0 to 1 of (4x+6)/((x+1)^2) is 4ln(2)+1