Mister Exam

Lang:

Other calculators

How to use it?
Integral of d{x}:
Integral of dy/e^y
Integral of e^x/(1+x^2)
Integral of 1/(x*(lnx)^2)
Integral of x*arctan(x)
Identical expressions
(y- one)/(sqrty+ one)
(y minus 1) divide by ( square root of y plus 1)
(y minus one) divide by ( square root of y plus one)
(y-1)/(√y+1)
y-1/sqrty+1
(y-1) divide by (sqrty+1)
Similar expressions
(y+1)/(sqrty+1)
(y-1)/(sqrty-1)

Integral of (y-1)/(sqrty+1) dx

The solution

You have entered [src]

  9             
  /             
 |              
 |    y - 1     
 |  --------- dy
 |    ___       
 |  \/ y  + 1   
 |              
/               
y

\int\limits_{y}^{9} \frac{y - 1}{\sqrt{y} + 1}\, dy

Integral((y - 1)/(sqrt(y) + 1), (y, y, 9))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Let $u = \sqrt{y}$ .
  
  Then let $du = \frac{dy}{2 \sqrt{y}}$ and substitute $du$ :
  
  $\int \left(2 u^{2} - 2 u\right)\, du$
  1. Integrate term-by-term:
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 2 u^{2}\, du = 2 \int u^{2}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{2}\, du = \frac{u^{3}}{3}$
      
      So, the result is: $\frac{2 u^{3}}{3}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- 2 u\right)\, du = - 2 \int u\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u\, du = \frac{u^{2}}{2}$
      
      So, the result is: $- u^{2}$
    The result is: $\frac{2 u^{3}}{3} - u^{2}$
  Now substitute $u$ back in:
  
  $\frac{2 y^{\frac{3}{2}}}{3} - y$
Method #2
1. Rewrite the integrand:
  
  $\frac{y - 1}{\sqrt{y} + 1} = \frac{y}{\sqrt{y} + 1} - \frac{1}{\sqrt{y} + 1}$
2. Integrate term-by-term:
  1. Let $u = \sqrt{y}$ .
    
    Then let $du = \frac{dy}{2 \sqrt{y}}$ and substitute $2 du$ :
    
    $\int \frac{2 u^{3}}{u + 1}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{u^{3}}{u + 1}\, du = 2 \int \frac{u^{3}}{u + 1}\, du$
      
      Rewrite the integrand:
      
      $\frac{u^{3}}{u + 1} = u^{2} - u + 1 - \frac{1}{u + 1}$
      
      Integrate term-by-term:
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{2}\, du = \frac{u^{3}}{3}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- u\right)\, du = - \int u\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u\, du = \frac{u^{2}}{2}$
      
      So, the result is: $- \frac{u^{2}}{2}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, du = u$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{1}{u + 1}\right)\, du = - \int \frac{1}{u + 1}\, du$
      
      Let $u = u + 1$ .
      
      Then let $du = du$ and substitute $du$ :
      
      $\int \frac{1}{u}\, du$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      Now substitute $u$ back in:
      
      $\log{\left(u + 1 \right)}$
      
      So, the result is: $- \log{\left(u + 1 \right)}$
      
      The result is: $\frac{u^{3}}{3} - \frac{u^{2}}{2} + u - \log{\left(u + 1 \right)}$
      
      So, the result is: $\frac{2 u^{3}}{3} - u^{2} + 2 u - 2 \log{\left(u + 1 \right)}$
    Now substitute $u$ back in:
    
    $\frac{2 y^{\frac{3}{2}}}{3} + 2 \sqrt{y} - y - 2 \log{\left(\sqrt{y} + 1 \right)}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{1}{\sqrt{y} + 1}\right)\, dy = - \int \frac{1}{\sqrt{y} + 1}\, dy$
    1. Let $u = \sqrt{y}$ .
      
      Then let $du = \frac{dy}{2 \sqrt{y}}$ and substitute $2 du$ :
      
      $\int \frac{2 u}{u + 1}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{u}{u + 1}\, du = 2 \int \frac{u}{u + 1}\, du$
      
      Rewrite the integrand:
      
      $\frac{u}{u + 1} = 1 - \frac{1}{u + 1}$
      
      Integrate term-by-term:
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, du = u$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{1}{u + 1}\right)\, du = - \int \frac{1}{u + 1}\, du$
      
      Let $u = u + 1$ .
      
      Then let $du = du$ and substitute $du$ :
      
      $\int \frac{1}{u}\, du$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      Now substitute $u$ back in:
      
      $\log{\left(u + 1 \right)}$
      
      So, the result is: $- \log{\left(u + 1 \right)}$
      
      The result is: $u - \log{\left(u + 1 \right)}$
      
      So, the result is: $2 u - 2 \log{\left(u + 1 \right)}$
      
      Now substitute $u$ back in:
      
      $2 \sqrt{y} - 2 \log{\left(\sqrt{y} + 1 \right)}$
    So, the result is: $- 2 \sqrt{y} + 2 \log{\left(\sqrt{y} + 1 \right)}$
  The result is: $\frac{2 y^{\frac{3}{2}}}{3} - y$
Add the constant of integration:

$\frac{2 y^{\frac{3}{2}}}{3} - y+ \mathrm{constant}$

The answer is:

\frac{2 y^{\frac{3}{2}}}{3} - y+ \mathrm{constant}

The answer (Indefinite) [src]

  /                             
 |                           3/2
 |   y - 1                2*y   
 | --------- dy = C - y + ------
 |   ___                    3   
 | \/ y  + 1                    
 |                              
/

\int \frac{y - 1}{\sqrt{y} + 1}\, dy = C + \frac{2 y^{\frac{3}{2}}}{3} - y

Simplify

The answer [src]

           3/2
        2*y   
9 + y - ------
          3

- \frac{2 y^{\frac{3}{2}}}{3} + y + 9

           3/2
        2*y   
9 + y - ------
          3

- \frac{2 y^{\frac{3}{2}}}{3} + y + 9

9 + y - 2*y^(3/2)/3

Use the examples entering the upper and lower limits of integration.

Other calculators

How to use it?

Identical expressions

Similar expressions

Integral of (y-1)/(sqrty+1) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2