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- Improper integral
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How to use it?
- Integral of d{x}:
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Integral of e^(2*x+1)
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Integral of (1+x)/x
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Integral of -15x^2
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Integral of (1+x^4)^(1/2)
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Identical expressions
- ln(x+ one)/(x+ one)²
- ln(x plus 1) divide by (x plus 1)²
- ln(x plus one) divide by (x plus one)²
- lnx+1/x+1²
- ln(x+1) divide by (x+1)²
- ln(x+1)/(x+1)²dx
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Similar expressions
- ln(x+1)/(x-1)²
- ln(x-1)/(x+1)²
Integral of ln(x+1)/(x+1)² dx
The solution
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[src]
2 / | | log(x + 1) | ---------- dx | 2 | (x + 1) | / 1
$$\int\limits_{1}^{2} \frac{\log{\left(x + 1 \right)}}{\left(x + 1\right)^{2}}\, dx$$
Integral(log(x + 1)/(x + 1)^2, (x, 1, 2))
The answer (Indefinite)
[src]
/ | | log(x + 1) 1 log(x + 1) | ---------- dx = C - ----- - ---------- | 2 x + 1 x + 1 | (x + 1) | /
$$\int \frac{\log{\left(x + 1 \right)}}{\left(x + 1\right)^{2}}\, dx = C - \frac{\log{\left(x + 1 \right)}}{x + 1} - \frac{1}{x + 1}$$
The graph
The answer
[src]
1 log(2) log(3) - + ------ - ------ 6 2 3
$$- \frac{\log{\left(3 \right)}}{3} + \frac{1}{6} + \frac{\log{\left(2 \right)}}{2}$$
=
=
1 log(2) log(3) - + ------ - ------ 6 2 3
$$- \frac{\log{\left(3 \right)}}{3} + \frac{1}{6} + \frac{\log{\left(2 \right)}}{2}$$
1/6 + log(2)/2 - log(3)/3
Use the examples entering the upper and lower limits of integration.