Question

Please help in integral calculus

Answer 1

-2x(5x²-6)

Explanation:

hope this helps you

Answer 2

First, express the fraction in partial fractions

Write it out as an identity:

`(2x-10)/(x(x+1)(x-1)) \equiv (A)/(x)+(B)/((x+1))+(C)/((x-1))`

Add the partial fractions:

`(2x-10)/(x(x+1)(x-1)) \equiv (A(x+1)(x-1)+Bx(x-1)+Cx(x+1))/(x(x+1)(x-1))`

Cancel the denominators from both sides of the original identity, so the numerators are equal:

`2x-10 \equiv A(x+1)(x-1)+Bx(x-1)+Cx(x+1)`

Now solve for A, B and C by substitution.

Substitute values of x which make one of the expressions equal zero (to eliminate all but one of A, B and C):

`\begin{aligned}x=0 \implies 2(0)-10 & =A(0+1)(0-1)+B(0)(0-1)+C(0)(0+1)\\-10 & = -A\\\implies A & = 10\end{aligned}`

`\begin{aligned}x=1 \implies 2(1)-10 & =A(1+1)(1-1)+B(1)(1-1)+C(1)(1+1)\\-8 & = 2C\\\implies C & = -4\end{aligned}`

`\begin{aligned}x=-1 \implies 2(-1)-10 & =A(-1+1)(-1-1)+B(-1)(-1-1)+C(-1)(-1+1)\\-12 & = 2B\\\implies B & = -6\end{aligned}`

Replace the found values of A, B and C in the original identity:

`\implies (2x-10)/(x(x+1)(x-1)) \equiv (10)/(x)-(6)/((x+1))-(4)/((x-1))`

Now integrate:

`\begin{aligned}\displaystyle \int (2x-10)/(x(x+1)(x-1))\:dx & =\int (10)/(x)-(6)/((x+1))-(4)/((x-1))\:\:dx\\\\& =10\int (1)/(x)\:dx\:\:-6 \int(1)/((x+1))\:dx\:\:-4 \int (1)/((x-1))\:dx\\\\& = 10 \ln |x|-6 \ln |x+1|-4 \ln |x-1|+C\end{aligned}`