How to solve this integral ?

int x^2/(x^4 + x^2 -2) dx∫x2x4+x2−2dx

2 Answers
Noah G
Jan 20, 2018

I = 1/6ln|x - 1| - 1/6ln|x + 1| + sqrt(2)arctan(x/sqrt(2))/3 + CI=16ln|x−1|−16ln|x+1|+√2arctan(x√2)3+C

Explanation:

In the denominator, letting y=x^2y=x2, we can rewrite as

y^2 + y -2y2+y−2

(y+ 2)(y - 1)(y+2)(y−1)

Thus,

(x^2 + 2)(x^2 - 1)(x2+2)(x2−1)

(x^2+ 2)(x + 1)(x - 1)(x2+2)(x+1)(x−1)

Now we can integrate using partial fractions:

A/(x + 1) + B/(x - 1) + C/(x^2 + 2) = x^2/((x^2 + 2)(x + 1)(x - 1))Ax+1+Bx−1+Cx2+2=x2(x2+2)(x+1)(x−1)

A(x - 1)(x^2 + 2) + B(x + 1)(x^2 + 2) + C(x + 1)(x - 1) = x^2A(x−1)(x2+2)+B(x+1)(x2+2)+C(x+1)(x−1)=x2

A(x^3 - x^2 + 2x - 2) + B(x^3 + x^2 + 2x + 2) + C(x^2 - 1) = x^2A(x3−x2+2x−2)+B(x3+x2+2x+2)+C(x2−1)=x2

Ax^3 - Ax^2 + 2Ax - 2A + Bx^3 + Bx^2 + 2Bx + 2B + Cx^2 - C = x^2Ax3−Ax2+2Ax−2A+Bx3+Bx2+2Bx+2B+Cx2−C=x2

Therefore,

{(A + B = 0), (B - A + C = 1), (2A + 2B = 0), (-2A+ 2B - C = 0):}

The first and third equations are the same. Therefore our new system becomes

{(A + B = 0), (B - A + C = 1), (2B - 2A - C = 0):}

We can add equation two to equation three eliminating variable C.

{(A+ B = 0), (3B - 3A = 1):}

You can easily solve this by elimination.

{(3A + 3B = 0), (-3A + 3B = 1):}

Therefore, 6B = 1 and B = 1/6. It follows that A = -1/6, and that C = 1- 1/3 = 2/3

Hence the integral becomes

I = int -1/(6(x + 1)) + 1/(6(x - 1)) + 2/(3(x^2 + 2))

The first and second terms are easily integrated. The third can be rearranged to the form 1/(x^2 + 1) and hence integrated using the arctangent identity.

I = 1/6ln|x - 1| - 1/6ln|x +1| + sqrt(2)arctan(x/sqrt(2))/3 + C

If you differentiate this you will get the original integral.

Hopefully this helps!

Ratnaker Mehta
Jan 20, 2018

sqrt2/3arc tan(x/sqrt2)+1/6ln|(x-1)/(x+1)|+C.

Explanation:

Letting x^2=y, we see that, the Integrand is,

y/(y^2+y-2)=y/((y+2)(y-1))=A/(y+2)+B/(y-1)," say, then",

by Heaviside's Method,

A=[y/(y-1)]_(y=-2)=(-2)/(-2-1)=2/3, and,

B=[y/(y+2)]_(y=1)=1/(1+2)=1/3.

:. y/(y^2+y-2)=y/((y+2)(y-1))=(2/3)/(y+2)+(1/3)/(y-1).

Replacing y" by "x^2, we have,

x^2/(x^4+x^2-2)=(2/3)/(x^2+2)+(1/3)/(x^2-1).

:. intx^2/(x^4+x^2-2)dx=2/3int1/(x^2+2)+1/3int1/(x^2-1)dx,

=2/3*1/sqrt2arc tan(x/sqrt2)+1/3*1/(2*1)ln|(x-1)/(x+1)|,

=sqrt2/3arc tan(x/sqrt2)+1/6ln|(x-1)/(x+1)|+C, as respected

HSBC244 has already derived!.

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