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  • Identical expressions

  • sqrt(x)*(lnx^ two)*x
  • square root of (x) multiply by (lnx squared ) multiply by x
  • square root of (x) multiply by (lnx to the power of two) multiply by x
  • √(x)*(lnx^2)*x
  • sqrt(x)*(lnx2)*x
  • sqrtx*lnx2*x
  • sqrt(x)*(lnx²)*x
  • sqrt(x)*(lnx to the power of 2)*x
  • sqrt(x)(lnx^2)x
  • sqrt(x)(lnx2)x
  • sqrtxlnx2x
  • sqrtxlnx^2x
  • sqrt(x)*(lnx^2)*xdx
Solving integrals/ sqrt(x)*(lnx^2)*x

Integral of sqrt(x)*(lnx^2)*x dx

Limits of integration:

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The solution

You have entered [src] 
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∫01xxlog⁡(x)2 dx\int\limits_{0}^{1} \sqrt{x} x \log{\left(x \right)}^{2}\, dx 
Integral(sqrt(x)*log(x)^2*x, (x, 0, 1))
Detail solution

  1. Let u=log⁡(x)u = \log{\left(x \right)}.

    Then let du=dxxdu = \frac{dx}{x} and substitute dudu:

    ∫u2e5u2 du\int u^{2} e^{\frac{5 u}{2}}\, du

    1. Use integration by parts:

      ∫u⁡dv⁡=u⁡v⁡−∫v⁡du⁡\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}

      Let u(u)=u2u{\left(u \right)} = u^{2} and let dv⁡(u)=e5u2\operatorname{dv}{\left(u \right)} = e^{\frac{5 u}{2}}.

      Then du⁡(u)=2u\operatorname{du}{\left(u \right)} = 2 u.

      To find v(u)v{\left(u \right)}:

      1. There are multiple ways to do this integral.

        Method #1

        1. Let u=5u2u = \frac{5 u}{2}.

          Then let du=5du2du = \frac{5 du}{2} and substitute 2du5\frac{2 du}{5}:

          ∫4eu25 du\int \frac{4 e^{u}}{25}\, du

          1. The integral of a constant times a function is the constant times the integral of the function:

            ∫2eu5 du=2∫eu du5\int \frac{2 e^{u}}{5}\, du = \frac{2 \int e^{u}\, du}{5}

            1. The integral of the exponential function is itself.

              ∫eu du=eu\int e^{u}\, du = e^{u}

            So, the result is: 2eu5\frac{2 e^{u}}{5}

          Now substitute uu back in:

          2e5u25\frac{2 e^{\frac{5 u}{2}}}{5}

        Method #2

        1. Let u=e5u2u = e^{\frac{5 u}{2}}.

          Then let du=5e5u2du2du = \frac{5 e^{\frac{5 u}{2}} du}{2} and substitute 2du5\frac{2 du}{5}:

          ∫425 du\int \frac{4}{25}\, du

          1. The integral of a constant times a function is the constant times the integral of the function:

            ∫25 du=2∫1 du5\int \frac{2}{5}\, du = \frac{2 \int 1\, du}{5}

            1. The integral of a constant is the constant times the variable of integration:

              ∫1 du=u\int 1\, du = u

            So, the result is: 2u5\frac{2 u}{5}

          Now substitute uu back in:

          2e5u25\frac{2 e^{\frac{5 u}{2}}}{5}

      Now evaluate the sub-integral.

    2. Use integration by parts:

      ∫u⁡dv⁡=u⁡v⁡−∫v⁡du⁡\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}

      Let u(u)=4u5u{\left(u \right)} = \frac{4 u}{5} and let dv⁡(u)=e5u2\operatorname{dv}{\left(u \right)} = e^{\frac{5 u}{2}}.

      Then du⁡(u)=45\operatorname{du}{\left(u \right)} = \frac{4}{5}.

      To find v(u)v{\left(u \right)}:

      1. Let u=5u2u = \frac{5 u}{2}.

        Then let du=5du2du = \frac{5 du}{2} and substitute 2du5\frac{2 du}{5}:

        ∫4eu25 du\int \frac{4 e^{u}}{25}\, du

        1. The integral of a constant times a function is the constant times the integral of the function:

          ∫2eu5 du=2∫eu du5\int \frac{2 e^{u}}{5}\, du = \frac{2 \int e^{u}\, du}{5}

          1. The integral of the exponential function is itself.

            ∫eu du=eu\int e^{u}\, du = e^{u}

          So, the result is: 2eu5\frac{2 e^{u}}{5}

        Now substitute uu back in:

        2e5u25\frac{2 e^{\frac{5 u}{2}}}{5}

      Now evaluate the sub-integral.

    3. The integral of a constant times a function is the constant times the integral of the function:

      ∫8e5u225 du=8∫e5u2 du25\int \frac{8 e^{\frac{5 u}{2}}}{25}\, du = \frac{8 \int e^{\frac{5 u}{2}}\, du}{25}

      1. Let u=5u2u = \frac{5 u}{2}.

        Then let du=5du2du = \frac{5 du}{2} and substitute 2du5\frac{2 du}{5}:

        ∫4eu25 du\int \frac{4 e^{u}}{25}\, du

        1. The integral of a constant times a function is the constant times the integral of the function:

          ∫2eu5 du=2∫eu du5\int \frac{2 e^{u}}{5}\, du = \frac{2 \int e^{u}\, du}{5}

          1. The integral of the exponential function is itself.

            ∫eu du=eu\int e^{u}\, du = e^{u}

          So, the result is: 2eu5\frac{2 e^{u}}{5}

        Now substitute uu back in:

        2e5u25\frac{2 e^{\frac{5 u}{2}}}{5}

      So, the result is: 16e5u2125\frac{16 e^{\frac{5 u}{2}}}{125}

    Now substitute uu back in:

    2x52log⁡(x)25−8x52log⁡(x)25+16x52125\frac{2 x^{\frac{5}{2}} \log{\left(x \right)}^{2}}{5} - \frac{8 x^{\frac{5}{2}} \log{\left(x \right)}}{25} + \frac{16 x^{\frac{5}{2}}}{125}

  2. Now simplify:

    2x52⋅(25log⁡(x)2−20log⁡(x)+8)125\frac{2 x^{\frac{5}{2}} \cdot \left(25 \log{\left(x \right)}^{2} - 20 \log{\left(x \right)} + 8\right)}{125}

  3. Add the constant of integration:

    2x52⋅(25log⁡(x)2−20log⁡(x)+8)125+constant\frac{2 x^{\frac{5}{2}} \cdot \left(25 \log{\left(x \right)}^{2} - 20 \log{\left(x \right)} + 8\right)}{125}+ \mathrm{constant}

The answer is:

2x52⋅(25log⁡(x)2−20log⁡(x)+8)125+constant\frac{2 x^{\frac{5}{2}} \cdot \left(25 \log{\left(x \right)}^{2} - 20 \log{\left(x \right)} + 8\right)}{125}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                                                 
 |                              5/2      5/2             5/2    2   
 |   ___    2               16*x      8*x   *log(x)   2*x   *log (x)
 | \/ x *log (x)*x dx = C + ------- - ------------- + --------------
 |                            125           25              5       
/
8 x52 (25 (log⁡x)24−5 log⁡x+2)125{{8\,x^{{{5}\over{2}}}\,\left({{25\,\left(\log x\right)^2}\over{4}} -5\,\log x+2\right)}\over{125}}
Simplify
The answer [src] 
 16
---
125
16125{{16}\over{125}} 
=
= 
 16
---
125
16125\frac{16}{125}
Упростить
Numerical answer [src] 
0.128
0.128

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

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