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- Integral of d{x}:
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Integral of e^(2*x+1)
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Integral of sin²xcos²x
- Integral of log(1+x)/(1+x^2)
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Integral of (4x-x^2)dx
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Identical expressions
- one /(x*(ln^2x+ sixteen))
- 1 divide by (x multiply by (ln squared x plus 16))
- one divide by (x multiply by (ln squared x plus sixteen))
- 1/(x*(ln2x+16))
- 1/x*ln2x+16
- 1/(x*(ln²x+16))
- 1/(x*(ln to the power of 2x+16))
- 1/(x(ln^2x+16))
- 1/(x(ln2x+16))
- 1/xln2x+16
- 1/xln^2x+16
- 1 divide by (x*(ln^2x+16))
- 1/(x*(ln^2x+16))dx
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Similar expressions
- 1/(x*(ln^2x-16))
Integral of 1/(x*(ln^2x+16)) dx
The solution
You have entered
[src]
1 / | | 1 | ---------------- dx | / 2 \ | x*\log (x) + 16/ | / 0
$$\int\limits_{0}^{1} \frac{1}{x \left(\log{\left(x \right)}^{2} + 16\right)}\, dx$$
Integral(1/(x*(log(x)^2 + 16)), (x, 0, 1))
The answer (Indefinite)
[src]
/ | | 1 / 2 \ | ---------------- dx = C + RootSum\64*z + 1, i -> i*log(32*i + log(x))/ | / 2 \ | x*\log (x) + 16/ | /
$$\int \frac{1}{x \left(\log{\left(x \right)}^{2} + 16\right)}\, dx = C + \operatorname{RootSum} {\left(64 z^{2} + 1, \left( i \mapsto i \log{\left(32 i + \log{\left(x \right)} \right)} \right)\right)}$$
The graph
The answer
[src]
1 / | | 1 | ---------------- dx | / 2 \ | x*\16 + log (x)/ | / 0
$$\int\limits_{0}^{1} \frac{1}{x \left(\log{\left(x \right)}^{2} + 16\right)}\, dx$$
=
=
1 / | | 1 | ---------------- dx | / 2 \ | x*\16 + log (x)/ | / 0
$$\int\limits_{0}^{1} \frac{1}{x \left(\log{\left(x \right)}^{2} + 16\right)}\, dx$$
Integral(1/(x*(16 + log(x)^2)), (x, 0, 1))
Use the examples entering the upper and lower limits of integration.