Find $$\int \frac{e^{x}(2-x^2)}{(1-x)\sqrt{1-x^2}}\mathrm dx$$

Looking at the numerator, combined with the surd, you can get $$\int\frac{e^x(1+x)\left(\frac{1}{\sqrt{1-x^2}}+\sqrt{1-x^2}\right)}{1-x^2}\mathrm dx$$Then this begins to look like the quotient rule, since the denominator is $\sqrt{1-x^2}^2$, and in the numerator, $1/\sqrt{1-x^2}$ is almost like the derivative of the square root. However, it isn't quite that - there are extra terms. $$\color{red}{e^x\left(\sqrt{1-x^2}+\frac{x}{\sqrt{1-x^2}}\right)}+\frac{e^x}{\sqrt{1-x^2}}+e^x x\sqrt{1-x^2}$$ The red terms are accounted for by quotient rule (giving an integral of $\frac{e^x}{\sqrt{1-x^2}}$). But then what do we do with the remaining terms?



Observe that the exponent of $1-x$ is $-\dfrac32$

So, let us find $$\dfrac{d\left(e^x\dfrac{(1+x)^n}{\sqrt{1-x}}\right)}{dx}$$

Compare with the given expression to find the value of $n$

  • $\begingroup$ Wow, that was surprisingly simple... With $n=1/2$, I suppose this means my method with the quotient rule was bound to fail! $\endgroup$ – John Doe Apr 4 at 1:55



where $f(x)=\dfrac{\sqrt{1+x}}{\sqrt{1-x}},$


Recall $\dfrac{d(e^xf(x))}{dx}=?$



$$-I=\int\dfrac{e^{\cos2t}(1+\sin^22t)}{\sin^2t}=e^{\cos2t}\csc^2t-\dfrac{d(e^{\cos2t})}{dt}(-\cot t)$$

$$=\dfrac{d(e^{\cos2t}(-\cot t))}{dt}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.