How do you integrate $\int_{0}^{1} \frac{1}{1 + {(x)}^{\frac{1}{2}}} d x$ $int_0^1 1/(1+(x)^(1/2)) dx$ ?

Question

How do you integrate $\int_{0}^{1} \frac{1}{1 + {(x)}^{\frac{1}{2}}} d x$ $int_0^1 1/(1+(x)^(1/2)) dx$ ?

2 Answers

Impact of this question

7221 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-14T04:04:11

Write $x^{\frac{1}{2}}$ $x^(1/2)$ as $\sqrt{x}$ $sqrtx$ :

$\int_{0}^{1} \frac{1}{1 + \sqrt{x}} d x$ $int_0^1 1/(1+sqrtx) dx$

To do the indefinite integral, I used a tip from WolframAlpha :

Let $u = \sqrt{x}$ $u = sqrtx$ then $d u = \frac{1}{2 \sqrt{x}} d x$ $du = 1/(2sqrtx)dx$

We need to solve for $d x$ $dx$ :

$d x = 2 \sqrt{x} d u$ $dx = 2sqrtxdu$

But $\sqrt{x} = u$ $sqrtx = u$ :

$d x = 2 u d u$ $dx = 2udu$

Perform the substitutions:

$\int \frac{1}{1 + {(x)}^{\frac{1}{2}}} d x = 2 \int \frac{u}{u + 1} d u$ $int 1/(1+(x)^(1/2)) dx = 2 int u/(u+1) du$

Add 0 to the numerator in the form if $+ 1 - 1$ $+1 - 1$

$\int \frac{1}{1 + {(x)}^{\frac{1}{2}}} d x = 2 \int \frac{u + 1 - 1}{u + 1} d u$ $int 1/(1+(x)^(1/2)) dx = 2 int (u+1-1)/(u+1) du$

Separate into two fractions:

$\int \frac{1}{1 + {(x)}^{\frac{1}{2}}} d x = 2 \int \frac{u + 1}{u + 1} - \frac{1}{u + 1} d u$ $int 1/(1+(x)^(1/2)) dx = 2 int (u+1)/(u+1)-1/(u+1) du$

The first fraction becomes 1:

$\int \frac{1}{1 + {(x)}^{\frac{1}{2}}} d x = 2 \int 1 - \frac{1}{u + 1} d u$ $int 1/(1+(x)^(1/2)) dx = 2 int 1-1/(u+1) du$

Both terms are trival to integrate:

$\int \frac{1}{1 + {(x)}^{\frac{1}{2}}} d x = 2 (u - ln (u + 1)) + C$ $int 1/(1+(x)^(1/2)) dx = 2 (u-ln(u+1))+C$

Reverse the substitution:

$\int \frac{1}{1 + {(x)}^{\frac{1}{2}}} d x = 2 (\sqrt{x} - ln (\sqrt{x} + 1)) + C$ $int 1/(1+(x)^(1/2)) dx = 2 (sqrtx-ln(sqrtx+1))+C$

To integrate from 0 to 1 we evaluate the right side at $x = 1$ $x = 1$ and subtract the right side evaluated at $x = 0$ $x = 0$ :

$\int_{0}^{1} \frac{1}{1 + {(x)}^{\frac{1}{2}}} d x = {2 (\sqrt{1} - ln (\sqrt{1} + 1)) + C} - {2 (\sqrt{0} - ln (\sqrt{0} + 1)) + C}$ $int_0^1 1/(1+(x)^(1/2)) dx = {2(sqrt1-ln(sqrt1+1))+C}-{2(sqrt0-ln(sqrt0+1))+C}$

The constants of integration sum to 0 and $\sqrt{0} - ln (1) = 0$ $sqrt0 - ln(1) = 0$ , therefore, the answer is:

$\int_{0}^{1} \frac{1}{1 + {(x)}^{\frac{1}{2}}} d x = 2 - 2 ln (2)$ $int_0^1 1/(1+(x)^(1/2)) dx = 2-2ln(2)$

Answer 2 · 2018-02-14T04:07:23

$f (x) = \frac{1}{1 + \sqrt{x}}$ $f(x) = 1/(1+sqrtx)$

let $\sqrt{x} = t$ $sqrtx=t$
$x = t^{2}$ $x=t^2$
differentiating both the sides,
$d x = 2 t . d t$ $dx = 2t.dt$

$\int f (x) = \int \frac{1}{1 + \sqrt{x}} d x$ $intf(x) = int1/(1+sqrtx) dx$
$= \int \frac{1}{1 + t} 2 t . d t = \int \frac{2 t}{1 + t} d t$ $= int1/(1+t) 2t. dt = int(2t)/(1+t) dt$
applying simple polynomial division,
$\int (2 - \frac{2}{1 + t}) d t$ $int (2-(2)/(1+t)) dt$
$\int 2 . d t - \int \frac{2}{1 + t} d t$ $int 2.dt-int(2)/(1+t) dt$
$2 \int d t - 2 \int \frac{1}{1 + t} d t$ $2intdt-2int1/(1+t) dt$
$2 t - 2 ln (1 + t) + c$ $2t-2ln(1+t) +c$

replacing, $t = \sqrt{x}$ $t=sqrtx$
$2 \sqrt{x} - 2 ln (1 + \sqrt{x}) + c$ $2sqrtx-2ln(1+sqrtx) +c$

now, applying the limits,
$2 \sqrt{1} - 2 ln (1 + \sqrt{1}) - (2 \sqrt{0} - 2 ln (1 + \sqrt{0}))$ $2sqrt1-2ln(1+sqrt1) - (2sqrt0-2ln(1+sqrt0))$
$2 - 2 ln (1 + 1) - (- 2 ln (1))$ $2-2ln(1+1) - (-2ln(1))$ as $\sqrt{1} = 1; \sqrt{0} = 0; ln 1 = 0$ $color(purple)(sqrt1=1; sqrt0=0;ln1=0$
$2 - 2 ln 2$ $2-2ln2$

Science

Math

Humanities

... and beyond

How do you integrate $\int_{0}^{1} \frac{1}{1 + {(x)}^{\frac{1}{2}}} d x$ $int_0^1 1/(1+(x)^(1/2)) dx$ ?

2 Answers

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you integrate int_0^1 1/(1+(x)^(1/2)) dx∫1011+(x)12dxint_0^1 1/(1+(x)^(1/2)) dx ?

2 Answers

Related questions

Impact of this question

How do you integrate $\int_{0}^{1} \frac{1}{1 + {(x)}^{\frac{1}{2}}} d x$ $int_0^1 1/(1+(x)^(1/2)) dx$ ?