Use the specified substitutions

With $u=e^x$, $du/dx=u$ and $dx/du=1/u$. The integral becomes $\int\dfrac1{u+1/u}\cdot\dfrac1u\,du=\int\dfrac1{u^2+1}\,du=\tan^{-1}u+C=\tan^{-1}(e^x)+C$.

With $u=2x-1$, $dx/du=1/2$ and $x=(u+1)/2$. Thus the integral is $\int\dfrac{(u+1)/2+1}{u^2}\cdot\dfrac12\,du=\dfrac14\int(1/u+3/u^2)\,du$. It equals $\dfrac14\ln|u|-\dfrac3{4u}+C=\dfrac14\ln|2x-1|-\dfrac3{4(2x-1)}+C$.

With $u=\sqrt x$, $u^2=x$ and $dx/du=2u$. Then $\int\dfrac1{2+u}\cdot2u\,du=2\int(1-2/(2+u))\,du$. This gives $2u-4\ln|2+u|+C=2\sqrt x-4\ln(2+\sqrt x)+C$. The logarithm argument is positive on x≥0; the substitution calculation applies first on x>0.

With $u=\cos x$, $du/dx=-\sin x$. The transformed integrand is $-\sin^2x=\cos^2x-1=u^2-1$. Integrate to get $u^3/3-u+C=\cos^3x/3-\cos x+C$. The final expression differentiates correctly even at points where the substitution division by sin x was only locally valid.

Take the principal substitution $x=\sin\theta$, with $-\pi/2<\theta<\pi/2$, so $dx=\cos\theta\,d\theta$ and $\sqrt{1-x^2}=\cos\theta>0$. The integral becomes $\int[\sin\theta\cos\theta]/\sqrt{\cos^2\theta}\,d\theta=\int\sin\theta\,d\theta=-\cos\theta+C=-\sqrt{1-x^2}+C$. The source triangle has hypotenuse 1, opposite side x and adjacent side √(1−x²). For negative x the signed principal-angle formulas, rather than a negative side length, justify the same result.

With $x=\tan\theta$ on $-\pi/2<\theta<\pi/2$, $dx=\sec^2\theta\,d\theta$. The integral becomes $\int\sec^2\theta/\sec^4\theta\,d\theta=\int\cos^2\theta\,d\theta=\dfrac12\int(1+\cos2\theta)\,d\theta$. This is $\dfrac12\sin\theta\cos\theta+\dfrac12\theta+C$. The source triangle has opposite side x, adjacent side 1 and hypotenuse √(1+x²). Using $\sin\theta=x/\sqrt{1+x^2}$, $\cos\theta=1/\sqrt{1+x^2}$ and $\theta=\tan^{-1}x$ gives $\dfrac{x}{2(1+x^2)}+\dfrac12\tan^{-1}x+C$.