Is it possible to integrate lnx/ (1+x²)?

 Is it possible to integrate lnx/ (1+x²)?

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Yes, it is possible to integrate $\frac{\ln 𝑥}{1+{𝑥}^2}$. Here is the step-by-step solution using integration by parts.

We start with the integral: $𝐼=\int \frac{\ln 𝑥}{1+{𝑥}^2}𝑑𝑥$

We use integration by parts, where we set $𝑢=\ln 𝑥$ and $𝑑𝑣=\frac{1}{1+{𝑥}^2}𝑑𝑥$. Then, we need to find $𝑑𝑢$ and $𝑣$: $𝑢=\ln 𝑥⟹𝑑𝑢=\frac{1}{𝑥}𝑑𝑥$ $𝑑𝑣=\frac{1}{1+{𝑥}^2}𝑑𝑥⟹𝑣=\arctan 𝑥$

Using the integration by parts formula: $\int 𝑢𝑑𝑣=𝑢𝑣-\int 𝑣𝑑𝑢$

Substituting $𝑢$, $𝑑𝑢$, $𝑣$, and $𝑑𝑣$: $𝐼=\int \frac{\ln 𝑥}{1+{𝑥}^2}𝑑𝑥=\ln 𝑥\cdot \arctan 𝑥-\int \arctan 𝑥\cdot \frac{1}{𝑥}𝑑𝑥$

We now need to evaluate the remaining integral: $𝐽=\int \frac{\arctan 𝑥}{𝑥}𝑑𝑥$

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To integrate $𝐽$, we use a substitution method. Let: $𝑡=\arctan 𝑥⟹𝑥=\tan 𝑡⟹𝑑𝑥={\sec }^2𝑡𝑑𝑡$

Now substitute into the integral: $𝐽=\int \frac{𝑡}{\tan 𝑡}\cdot {\sec }^2𝑡𝑑𝑡$

Since $\tan 𝑡=𝑥$, we have: ${\sec }^2𝑡=1+{\tan }^2𝑡=1+{𝑥}^2$ Substituting $\tan 𝑡=𝑥$ and ${\sec }^2𝑡=1+{𝑥}^2$ back, we get: $𝐽=\int 𝑡𝑑𝑡$

The integral of $𝑡$ is straightforward: $𝐽=\int 𝑡𝑑𝑡=\frac{{𝑡}^2}{2}+𝐶$

Substituting back $𝑡=\arctan 𝑥$: $𝐽=\frac{(\arctan 𝑥{)}^2}{2}+𝐶$

Therefore, the integral $𝐼$ becomes: $𝐼=\ln 𝑥\cdot \arctan 𝑥-\int \arctan 𝑥\cdot \frac{1}{𝑥}𝑑𝑥$ $𝐼=\ln 𝑥\cdot \arctan 𝑥-\frac{(\arctan 𝑥{)}^2}{2}+𝐶$

So, the integral of $\frac{\ln 𝑥}{1+{𝑥}^2}$ is: $\int \frac{\ln 𝑥}{1+{𝑥}^2}𝑑𝑥=\ln 𝑥\cdot \arctan 𝑥-\frac{(\arctan 𝑥{)}^2}{2}+𝐶$