Mathematics Test - 26

It is given that, \(f(x)=\sqrt{x}\) and \( g(x)=e^{x}-1\)

Therefore, \(\operatorname{fog}(x)= f(g(x))=\sqrt{e^{x}-1}\)

Now, it is also given that,

\(\int \operatorname{fog}(x) d x=A \operatorname{fog}(x)+ B \operatorname{tan}^{-1}(\operatorname{fog}(x))+C\) ----(1)

Therefore,

\(\Rightarrow \int \operatorname{fog}(x)~ d x=\int \sqrt{e^{x}-1} ~d x\) ----(2)

Let \(t=\sqrt{e^{x}-1}\)

\(\Rightarrow t^2= e^x-1\)

Differentiating both sides, we get,

\(\Rightarrow 2t~dt=e^x~dx\)

\(\Rightarrow 2t~dt=t^2+1~dx \quad (\because t^2= e^x-1)\)

\(\Rightarrow \frac{2t}{t^2+1}~dt = dx\)

Substituting this value in equation (2), we get,

\(\Rightarrow\int t.\frac{2 t}{t^{2}+1} ~d t=\int \frac{2 t^{2}}{t^{2}+1} ~d t\)

\(\Rightarrow2\int \frac{ t^{2}+1-1}{t^{2}+1} ~d t\)

\(\Rightarrow2\int \frac{ t^{2}+1}{t^{2}+1} ~d t -2\int\frac{dt}{t^{2}+1}\)

\(\Rightarrow2\int d t -2\int\frac{dt}{t^{2}+1}\)

\(\Rightarrow2 t-2 \tan ^{-1} t+C\)

Putting \(t=\sqrt{e^{x}-1}\),

\(\Rightarrow2 \sqrt{e^{x}-1}-2 \tan ^{-1}\left(\sqrt{e^{x}-1}\right)+C\)

We can write this equation as,

\(=2 \operatorname{fog}(x)-2 \tan ^{-1}(\operatorname{fog}(x))+C\)

Comparing this equation with equation (1), we get \(A=2\) and \(B=-2\)

Therefore, \(A+B=2+(-2)=0\)