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Find the indefinite integral x divided by x plus 1 quantity squared dx?

Anonymous

∙ 14y ago

Updated: 11/2/2022

∫(x/(x+1)2)dx

=∫((x+1-1)/(x+1)2)dx

=∫(1/(x+1))dx - ∫(1/(x+1)2)dx

u=x+1, du=dx

∫(1/u)du - ∫(1/u2)du

=log(u) - (-1/u) + C

=log(x+1) + 1/(x+1) + C

Wiki User

∙ 14y ago

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What is the integral of 1 divided by x squared?

The indefinite integral of (1/x^2)*dx is -1/x+C.

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What is the definite integral of 1 divided by x squared?

For it to be a definite integral, you would need to specify a range. We can however give you the indefinite integral. The easiest way to do this is to think of it not as a fraction, but as a negative exponent: 1/x2 = x-2 It then becomes quite easy to integrate, as we can say in general: ∫(axn) dx = ax(n + 1) / (n + 1) + C In this case then, we have: ∫(x-2) dx = -x-1 + C, or -1/x + C

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