Question

Integration of x/x^2+y^2 with respect to x

Answer 1

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Answer 2

Answer:

$\int \frac{x}{x^{2}+y^{2}} d x=\frac{\ln \left(x^{2}+y^{2}\right)}{2}$

Step-by-step explanation:

Given:

$\int \frac{x}{x^{2}+y^{2}} d x$

To find:

the integration with respect to x.

Step 1

we are integrating with respect to x that's why y will be treated as a constant. When we differentiate the denominator, we will get the numerator value.

So we are substituting,

$x^{2}+y^{2}=z$

Taking derivative both sides of the equation, we get

$&d\left(x^{2}+y^{2}\right)=d z$

$&2 x d x=d z$

$&x d x=\frac{d z}{2}$

Step 2

Replacing x in the terms of z, then we get

$&\int \frac{x}{x^{2}+y^{2}} d x=\int \frac{1}{z \times 2} d z$

$&=\frac{1}{2} \int \frac{1}{z} d z$

By using the formula

$\int \frac{1}{a x+b} d x=\frac{\ln |a x+b|}{a}+c$

where a = 1 and b = 0.

$=\frac{1}{2} \ln |z|+c$

Where c exists an integral constant.

Step 3

Replacing z in the terms of x.

$\int \frac{x}{x^{2}+y^{2}} d x=\frac{\ln \left(x^{2}+y^{2}\right)}{2}$

Therefore,

$\int \frac{x}{x^{2}+y^{2}} d x=\frac{\ln \left(x^{2}+y^{2}\right)}{2}$

SPJ2