Question

Integrate
x^3+1/(sqrt(x^2+x))​

Answer 1

$3: 333 </p><p> x </p><p>Q How do i integrat...</p><p> \square \squareं quora.com</p><p>Evaluate</p><p> I=\int \frac{x-1}{(x+1) \sqrt{x^{3}+x^{2}+x}} d x </p><p>Rationalisation is the convenient way to start with</p><p></p><p> \Rightarrow I=\int \frac{(x-1)(x+1)}{(x+1)^{2} \sqrt{x^{3}+x^{2}+x}} d x </p><p> \Rightarrow I=\int \frac{x^{2}-1}{\left(x^{2}+1+2 x\right) \sqrt{x^{3}+x^{2}+x}} d x </p><p> \Rightarrow I=\int \frac{\left(1-\frac{1}{2}\right)}{\left(x+\frac{1}{x}+2\right) \sqrt{x+1+\frac{1}{x}} d x} d x </p><p>Thetititit</p><p> \left(x+1+\frac{1}{x}\right)=t </p><p> \Rightarrow\left(1-\frac{1}{x^{2}}\right) d x=d t </p><p> \Rightarrow I=\int \frac{1}{(t+1) \sqrt{t}} d t </p><p> \Rightarrow I=\int \frac{1+t-t}{(t+1) \sqrt{t}} d t </p><p> \Rightarrow I=\int \frac{1+t}{(t+1) \sqrt{t}} d t-\int \frac{t}{(t+1) \sqrt{t}} d t </p><p> \Rightarrow I=\int \frac{1}{\sqrt{t}} d t-\int \frac{\sqrt{t}}{t+1} d t </p><p> \Rightarrow I=2 \sqrt{t}-\underbrace{\int \frac{\sqrt{t}}{t+1} d t}_{I_{1}} </p><p>Let's look at I_{1} I_{1}=\int \frac{\sqrt{t}}{t+1} d t </p><p>Substitute: \sqrt{t}=u \Rightarrow \frac{1}{2 \sqrt{t}}^{t=d u} </p><p> \Rightarrow d t=2 \sqrt{\text { ftan }} </p><p> \Rightarrow d t=2 u d u </p><p> \Rightarrow I_{1}=2 \int \frac{u^{2}}{u^{2}+1} d u </p><p> \Rightarrow I_{1}=2 \int \frac{u^{2}+1-1}{u^{2}+1} d u$