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f(x)=(x^3)/((1+x)^(1/2)^4)−6−5−4−3−2−10123456−3-2.5−2-1.5−1-0.500.511.522.53xy

Exercise

∫(x3(1+x)4)dx\int\left(\frac{x^3}{\left(\sqrt{1+x}\right)^4}\right)dx

Step-by-step Solution

1

Simplify (1+x)4\left(\sqrt{1+x}\right)^4 using the power of a power property: (am)n=am⋅n\left(a^m\right)^n=a^{m\cdot n}. In the expression, mm equals 12\frac{1}{2} and nn equals 44

∫x3(1+x)2dx\int\frac{x^3}{\left(1+x\right)^{2}}dx
2

Expand

∫x31+2x+x2dx\int\frac{x^3}{1+2x+x^2}dx
3

Divide x3x^3 by 1+2x+x21+2x+x^2

;x2+2x;+1;;x;−2;;;x2+2x;+1);x3−;xn−;xn−;xn‾;x2+2x;+1;−x3−2x2−x;−;xn‾−x3−2x2−x;;−2x2−x;−;xn;x2+2x;+1−;xn;;2x2+4x;+2;;‾;;2x2+4x;+2;;−;xn;;3x;+2;;\begin{array}{l}\phantom{\phantom{;}x^{2}+2x\phantom{;}+1;}{\phantom{;}x\phantom{;}-2\phantom{;}\phantom{;}}\\\phantom{;}x^{2}+2x\phantom{;}+1\overline{\smash{)}\phantom{;}x^{3}\phantom{-;x^n}\phantom{-;x^n}\phantom{-;x^n}}\\\phantom{\phantom{;}x^{2}+2x\phantom{;}+1;}\underline{-x^{3}-2x^{2}-x\phantom{;}\phantom{-;x^n}}\\\phantom{-x^{3}-2x^{2}-x\phantom{;};}-2x^{2}-x\phantom{;}\phantom{-;x^n}\\\phantom{\phantom{;}x^{2}+2x\phantom{;}+1-;x^n;}\underline{\phantom{;}2x^{2}+4x\phantom{;}+2\phantom{;}\phantom{;}}\\\phantom{;\phantom{;}2x^{2}+4x\phantom{;}+2\phantom{;}\phantom{;}-;x^n;}\phantom{;}3x\phantom{;}+2\phantom{;}\phantom{;}\\\end{array}
4

Resulting polynomial

∫(x−2+31+x+−1(1+x)2)dx\int\left(x-2+\frac{3}{1+x}+\frac{-1}{\left(1+x\right)^{2}}\right)dx
5

Simplify the expression

∫xdx+∫−2dx+∫31+xdx+∫−1(1+x)2dx\int xdx+\int-2dx+\int\frac{3}{1+x}dx+\int\frac{-1}{\left(1+x\right)^{2}}dx
6

The integral ∫xdx\int xdx results in: 12x2\frac{1}{2}x^2

12x2\frac{1}{2}x^2
7

The integral ∫−2dx\int-2dx results in: −2x-2x

−2x-2x
8

The integral ∫31+xdx\int\frac{3}{1+x}dx results in: 3ln⁡(x+1)3\ln\left(x+1\right)

3ln⁡(x+1)3\ln\left(x+1\right)
9

The integral ∫−1(1+x)2dx\int\frac{-1}{\left(1+x\right)^{2}}dx results in: 11+x\frac{1}{1+x}

11+x\frac{1}{1+x}
10

Gather the results of all integrals

12x2−2x+3ln⁡∣x+1∣+11+x\frac{1}{2}x^2-2x+3\ln\left|x+1\right|+\frac{1}{1+x}
11

As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration CC

12x2−2x+3ln⁡∣x+1∣+11+x+C0\frac{1}{2}x^2-2x+3\ln\left|x+1\right|+\frac{1}{1+x}+C_0

Final answer to the exercise

12x2−2x+3ln⁡∣x+1∣+11+x+C0\frac{1}{2}x^2-2x+3\ln\left|x+1\right|+\frac{1}{1+x}+C_0

Try other ways to solve this exercise

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∫(x3(√1+x)4 )dx
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