User:Yorik sar

a)
$$ \int{\frac{\sin x}{\sqrt{\cos x}} dx} = \bigg\{ \sin x dx = -(\cos x)' dx = -d(\cos x) \bigg\} = - \int{\frac{d(\cos x)}{\sqrt{\cos x}}} = \bigg\{ y = \cos x \bigg\} = - \int{y^{-\frac 1 2}dy} = -2y^{\frac 1 2} + C = -2\sqrt{\cos x} + C $$

b)
$$ \int\ln(x + 3)dx = \bigg\{ dx = d(x+3) \bigg\} = \int\ln(x+3)d(x+3) = \bigg\{ v = x+3, u = \ln(x+3) = \ln v \bigg\} = \int u dv = $$$$ = u v - \int v du = v \ln v - \int v d(\ln v) = v \ln v - \int \frac v v dv = v \ln v - v + C = (x + 3) \ln (x + 3) - x + C $$

c)
$$ \int \frac {x^3+x^2-4x+1} {x^2+x-6} dx = \int \left( x + \frac {2x+1} {x^2+x-6} \right) dx = \int x dx + \int \frac {d(x^2+x-6)}{x^2+x-6} = \frac {x^2} 2 + \ln |x^2+x-6| + C $$

d)
$$ \int\limits_0^1 \arctan x dx = \left( x \arctan x \right) \bigg|_0^1 - \int\limits_0^1 xd(\arctan x) = \arctan 1 - \int\limits_0^1 \frac {xdx} {1+x^2} = \frac \pi 4 - \frac 1 2 \int\limits_0^1 \frac {d(1+x^2)} {1+x^2} = \frac \pi 4 - \frac 1 2 \left( \ln(1+x^2) \right) \bigg|_0^1 = \frac \pi 4 - \frac 1 2 \ln 2 $$